PORTUGALIAE MATHEMATICA
Vol. 55 Fasc. 3 – 1998
CHEBYSHEV POLYNOMIALS AND
SOME METHODS OF APPROXIMATION
Gheorghe Udrea
1 – Chebyshev polynomials (Tn (x))n≥0 and (Un (x))n≥0 of the first and the
second kind, respectively, are defined by the recurrence relations:
(1.1)
Tn+1 (x) = 2x · Tn (x) − Tn−1 (x) ,
x ∈ C, n ∈ N ∗ ,
with T0 (x) = 1, T1 (x) = x;
(1.2)
Un+1 (x) = 2x · Un (x) − Un−1 (x) ,
x ∈ C, n ∈ N ∗ ,
where U0 (x) = 1, U1 (x) = 2x.
en (x))n≥0 —
On the other hand there are the sequences (Ten (x))n≥0 and (U
“associated” of the Chebyshev polynomials (Tn (x))n≥0 and (Un (x))n≥0 , respectively — defined by
(1.3)
Ten+1 (x) = 2x · Ten (x) + Ten+1 (x) ,
with Te0 (x) = 1 and Te1 (x) = x;
(1.4)
en+1 (x) = 2x · U
en (x) + U
en−1 (x) ,
U
x ∈ C, n ∈ N ∗ ,
x ∈ C,
n ∈ N∗ ,
e0 (x) = 1, U
e1 (x) = 2x.
where U
en )n≥0 and the
There is a simple connection between the sequences (Ten )n≥0 , (U
sequences (Tn )n≥0 and (Un )n≥0 , respectively:
(1.5)

Tk (i · x)


e

,
 Tk (x) =
k
i

Te (i · x)


 Tk (x) = k
,
k
i
x ∈ C, k ∈ N ,
Received : November 12, 1996; Revised : April 29, 1997.
262
G. UDREA

Uk (i · x)


e

,
 Uk (x) =
k
i
(1.6)
e (i · x)

U


 Uk (x) = k
,
k
x ∈ C, k ∈ N ,
i
where i2 = −1.
Also, there is an interesting connection between the sequence (Fn )n≥0 of
Fibonacci numbers
(1.7)
Fn+1 = Fn + Fn−1 , n ∈ N∗ ,
F0 = 0, F1 = 1 ,
the sequence (Ln )n≥0 of Lucas numbers
(1.8)
Ln+1 = Ln + Ln−1 , n ∈ N∗ ,
L0 = 2, L1 = 1 ,
en )n≥0 :
and, on the other hand, the sequences (Tn )n≥0 , (Ten )n≥0 , (Un )n≥0 , (U
(1.9)
Tn
(1.90 )
Ten
(1.10)
µ ¶
i
2
=
1 n
· i · Ln ,
2
µ ¶
1
2
=
1
· Ln ,
2
Un
en
U
(1.100 )
µ ¶
i
2
µ ¶
1
2
n∈N,
n∈N,
= in · Fn+1 ,
= Fn+1 ,
n∈N,
n ∈ N, i2 = −1 .
One has, also, the remarkable formulas:
ϕ ∈ C, k ∈ N ,
(1.11)
Tk (cos ϕ) = cos k ϕ ,
(1.110 )
Tek (i · cos ϕ) = ik · cos kϕ ,
and
sin k ϕ
,
sin ϕ
(1.12)
Uk−1 (cos ϕ) =
(1.120 )
ek−1 (i · cos ϕ) = ik−1 ·
U
(see (1.5)–(1.6)).
ϕ ∈ C, k ∈ N ,
ϕ ∈ C, sin ϕ 6= 0, k ∈ N∗ ,
sin k ϕ
,
sin ϕ
ϕ ∈ C, sin ϕ 6= 0, k ∈ N∗ ,
CHEBYSHEV POLYNOMIALS
263
e m (a))m≥1 defined by
2 – Let us consider the sequence (λ
e m (a) =
λ
(2.1)
e
U
p(m+1)−1 (a)
epm−1 (a)
U
,
m ∈ N∗ ,
where a ∈ R∗ and p ∈ N∗ .
We wish now to find a quadratic equation to be satisfied by the limit of the
e m (a))m≥1 . In order to obtain this equation we need to prove the
sequence (λ
following
Lemma. If (Tn (x))n≥0 and (Un (x))n≥0 are the sequences of Chebyshev
polynomials of the first kind and the second kind, respectively, then one has
(2.2)
Um+p−1 (a) = 2 · Tp (a) · Um−1 (a) − Um−p−1 (a) ,
∀ m ≥ p + 1, m, p ∈ N∗ , ∀ a ∈ C.
Proof: Indeed, let a be an element of C; then ∃ ϕ ∈ C such that a = cos ϕ.
We have
Um+p−1 (a) + Um−p−1 (a) = Um+p−1 (cos ϕ) + Um−p−1 (cos ϕ) =
=
sin(m + p) ϕ + sin(m − p) ϕ
sin(m + p) ϕ sin(m − p) ϕ
+
=
sin ϕ
sin ϕ
sin ϕ
=
sin m ϕ
2 · sin m ϕ · cos p ϕ
= 2 · cos p ϕ ·
= 2 · Tp (cos ϕ) · Um−1 (cos ϕ)
sin ϕ
sin ϕ
= 2 · Tp (a) · Um−1 (a) ,
q.e.d. .
Now, from (2.2) and (1.5)–(1.6) we obtain
(2.3)
em+p−1 (a) = 2 · Tep (a) · U
em−1 (a) − (−1)p · U
em−p−1 (a) ,
U
∀ m ≥ p + 1, m, p ∈ N∗ , ∀ a ∈ C.
em−1 (a) and replacing m by mp, we find
Dividing equation (2.3) through by U
(2.4)
e m (a) = 2 · Tep (a) −
λ
(−1)p
e m−1 (a)
λ
,
m≥2.
e ∞ (a) = limm→∞ λ
e m (a), a ∈ R∗ , therefore satisfies the quadratic
The limit λ
equation
(2.5)
e ∞ (a))2 − 2 · Tep (a) · λ
e ∞ (a) + (−1)p = 0 .
(λ
264
G. UDREA
Remarks.
e ∞ (a) = limm→∞ λ
e m (a) can be obtained from (2.1) and
a) The limit λ
(2.6)
One obtains
en (a) =
U
³
a+
´n ³
´n
√
√
a2 + 1 − a − a2 + 1
√
,
2 · a2 + 1
³
´p
√

 a − a2 + 1 ,
e ∞ (a) = ³
λ
´p
√

 a + a2 + 1 ,
(2.7)
a∈R.
a<0,
a>0.
b) From (2.5), (1.110 ) and (1.120 ), we have the same result:
e ∞ (a) = Tep (a) ±
λ
p
= ip · Tp (b) ± i ·
³
ep−1 (a) = Tep (i · b) ±
a2 + 1 · U
p
q
p
a2 + 1
´p
´
³
p
b2 − 1 · Up−1 (b)
cos2 ϕ − 1 · Up−1 (cos ϕ) = ip · cos p ϕ ± i sin ϕ
³
³
= ip · (cos ϕ ± i sin ϕ)p = i · b ±
³
³
ep−1 (i · b) =
(i · b)2 + 1 · U
b2 − 1 · ip−1 · Up−1 (b) = ip · Tp (b) ±
= ip · Tp (cos ϕ) ±
= a±
q
,
q.e.d. .
p
b2 − 1
´´p
³
= i·b±
q
(i · b)2 + 1
´
sin p ϕ ´
sin ϕ
´p
Hence
e ∞ (a) =
λ
(2.70 )
(
√
ep−1 (a), a < 0 ,
a2 + 1 · U
√
ep−1 (a), a > 0 .
Tep (a) + a2 + 1 · U
Tep (a) −
Clearly we utilized, in b), the identity
em−1 (x))2 = (−1)m ,
(Tem (x))2 − (x2 + 1) · (U
(2.8)
From (2.5), we obtain, for a =
(θ =
√
1+ 5
2
(2.9)
(see [1]).
1
2,
∀ x ∈ C, ∀ m ∈ N ∗ .
that the number θ p =
1
2
· (Lp +
— the “golden ratio”) satisfies the quadratic equation
x2 − Lp · x + (−1)p = 0 ,
√
5 · Fp )
265
CHEBYSHEV POLYNOMIALS
3 – The Aitken sequence (x∗n )n≥2 is the sequence defined by
x∗n =
(3.1)
xn+1 · xn−1 − x2n
,
xn+1 − 2 · xn + xn−1
n≥2,
where (xn )n≥1 is a convergent sequence.
In this section we establish the result
e n+r (a) · λ
e n−r (a) − (λ
e n (a))2
λ
(3.2)
e n+r (a) − 2 · λ
e n (a) + λ
e n−r (a)
λ
e 2n (a) ,
=λ
e n (a))n≥1 is the sequence defined by (2.1) and a ∈ R∗ .
1 ≤ r < n, where (λ
i2
Proof: Let a be an element of C; then ∃ ϕ ∈ C such that a = i · cos ϕ,
= −1. One has
a)
e n+r (a)·λ
e n−r (a)−(λ
e n (a))2 = λ
e n+r (i·cos ϕ)·λ
e n−r (i·cos ϕ)−(λ
e n (i·cos ϕ))2 =
λ
=
e
e
U
p(n+r+1)−1 (i · cos ϕ) Up(n−r+1)−1 (i · cos ϕ)
·
−
e
e
U
U
p(n+r)−1 (i · cos ϕ)
p(n−r)−1 (i · cos ϕ)
= ip ·
µ
· cos ϕ)
ep n−1 (i · cos ϕ)
U
¶2
¶2
sin p ϕ · sin2 p r ϕ · sin p(2n + 1) ϕ
;
sin2 p n ϕ · sin p(n + 1) ϕ · sin p(n − r) ϕ
e n+r (a) − λ
e n (a) = λ
e n+r (i · cos ϕ) − λ
e n (i · cos ϕ)
λ
= ... = ip+1 ·
c)
p(n+1)−1 (i
sin p(n + 1) ϕ
sin p(n + r + 1) ϕ p sin p(n − r + 1) ϕ
·i ·
− ip ·
sin p(n + r) ϕ
sin p(n − r) ϕ
sin p n ϕ
= ... = (−1)p ·
b)
µe
U
sin p ϕ · sin p r ϕ
;
sin p n ϕ · sin p(n + r) ϕ
³
´
³
e n+r (a) − 2 · λ
e n (a) + λ
e n−r (a) = λ
e n+r (a) − λ
e n (a) − λ
e n (a) − λ
e n−r (a)
λ
= ... = ip ·
2 · sin p ϕ · sin2 p r ϕ · cos p n ϕ
.
sin p n ϕ · sin p(n − r) ϕ · sin p(n + r) ϕ
´
Finally, on combining a), b) and c), we derive the required result:
(∗)
e n+r (a) · λ
e n−r (a) − (λ
e n (a))2
λ
e n+r (a) − 2 · λ
e n (a) + λ
e n−r (a)
λ
e 2n (a) ,
=λ
1≤r<n.
e ∗ (a))n≥2 verifies the
For r = 1 in (∗) one obtains that the Aitken sequence (λ
n
relation
(3.3)
e ∗ (a) = λ
e 2n (a) ,
λ
n
n ≥ 2, a ∈ R∗ .
266
G. UDREA
e n (a))n≥1 , a ∈ R∗ , and applied Aitken ac4 – We began with a sequence (λ
e ∗ (a))n≥2 . We now investigate the application of
celeration to give a sequence (λ
n
∗
e
Aitken acceleration to (λn (a))n≥2 , a ∈ R∗ , and so on, repeatedly. It is helpful to
use a different notation. Let us write
(k)
e (k+1) (a) =
λ
n
(4.1)
(0)
(k)
³
(k)
e
e
e
λ
n+1 (a) · λn−1 (a) − λn (a)
(k)
(k)
(k)
´2
e
e
e
λ
n+1 (a) − 2 · λn (a) + λn−1 (a)
,
e n (a) = λ
e n (a), a ∈ R∗ , n ∈ N∗ .
for k = 0, 1, 2, 3, ..., where λ
(0)
e n (a))n≥1 is our original sequence and (λ
e (1)
Thus (λ
n (a))n≥2 is the sequence
e ∗ (a))n≥2 . For k ≥ 0, the (k + 1)th sequence
which we have hitherto called (λ
n
e (k+1)
(λ
(a))
is
obtained
by
using Aitken acceleration on the sequence
n
n≥k+1
(k)
e
(λn (a))n≥k .
We have already seen from (3.3) that
e (1) (a) = λ
e 2n (a) ,
λ
n
(4.2)
n ≥ 2, a ∈ R∗ .
It follows from this and (4.1) that
(1)
e (2) (a) =
λ
n
(1)
2
e
e
λ1n+1 (a) · λ
n−1 (a) − (λn (a))
(1)
(1)
e
e
e1
λ
n+1 (a) − 2 · λn (a) + λn−1 (a)
=
e 2n+1 (a) · λ
e 2n−1 (a) − (λ
e 2n (a))2
λ
e 2n+2 (a) − 2 λ
e 2n (a) + λ
e 2n−2 (a)
λ
.
Using (3.2) with n replaced by 2n and with r = 2, we deduce that
(4.3)
e
λ(2)
n (a) = λ4n (a) ,
n≥3.
Finally, it follows by induction that
(4.4)
e ∞ (a) ,
e (k) (a) = λ
e k (a) −→ λ
λ
n·2
n
k
which holds for each k ≥ 0 and all n ≥ k + 1.
5 – Let us consider now √
Newton’s method: given an initial approximation a0
e
to the number λ∞ (a) = (a+ a2 + 1)p , a ∈ R∗ , we compute the Newton sequence
(an )n≥0 from
an+1 = an −
(5.1)
=
a2n − 2 · Tep (a) · an + (−1)p
³
2 · an − Tep (a)
a2n − (−1)p
³
2 · an − Tep (a)
´,
´
n≥0.
267
CHEBYSHEV POLYNOMIALS
e (a) for some values of k ∈ N∗ , we find that
If an = λ
k
an+1 =
=
=
i.e.,
(5.2)
since:
a2n − (−1)p
³
2 · an − Tep (a)
´ =
e (a))2 − (−1)p
(λ
k
³
e (a) − Tep (a)
2· λ
k
´ = ...
2
p
2
e
e
(U
p(k+1)−1 (a)) − (−1) · (Upk−1 (a))
³
epk−1 (a) · U
e
e
e
2·U
p(k+1)−1 (a) − Tp (a) · Upk−1 (a)
e
U
p·(2k+1)−1 (a)
ep·2k−1 (a)
U
´
e (a) ,
=λ
2k
e (a) ,
an+1 = λ
2k
a ∈ R∗ ,
2
p
2
e
e
e
e
1) (U
p·(k+1)−1 (a)) − (−1) · (Upk−1 (a)) = Up−1 (a) · Up·(2k+1)−1 (a),
∀ a ∈ C, ∀ p ∈ N ∗ , ∀ k ∈ N ∗ ;
e
e
e
e
e
2) U
p·(2k+1)−1 (a) − Tp (a) · Upk−1 (a) = Up−1 (a) · Tpk (a),
∗
∗
∀ a ∈ C, ∀ p ∈ N , ∀ k ∈ N ;
e2m−1 (a) = 2 · Tem (a) · U
em−1 (a), ∀ a ∈ C, ∀ m ∈ N∗ .
3) U
e
U
(a)
e 1 (a) = 2p−1
= 2· Tep (a),
Thus, if we choose as the initial approximant a0 = λ
ep−1 (a)
U
a ∈ R∗ , we see by induction that
(5.20 )
e 2·1 (a) = λ
e 2 (a) ,
a1 = a0+1 = λ
(5.200 )
e 2·2 (a) ,
a2 = a1+1 = λ
(5.2000 )
e 2 (a) ,
a3 = a2+1 = λ
2·2
Hence
(5.3)
e 2 (a) ;
a1 = λ
e 2 (a) ;
a2 = λ
2
e 3 (a) .
a3 = λ
2
e 2n (a) −→ λ
e ∞ (a) .
an = λ
n
268
G. UDREA
6 – Finally, we consider the secant method, in which we approximate to a
root of the equation f (x) = 0 as follows: we choose two initials approximants a 1
and a2 and compute the sequence (an )n≥2 from
(6.1)
an+1 = an −
f (an ) · (an − an−1 )
,
f (an ) − f (an−1 )
n≥2.
e (a) and
In our case f (x) = x2 − 2 · Tep (a) · x + (−1)p and, if we choose a1 = λ
k
∗
∗
e
a2 = λm (a), a ∈ R , for some m, k ∈ N , m 6= k, we find that
a3 =
=
=
³
a1 · a2 − (−1)p
a1 · f (a2 ) − a2 · f (a1 )
=
= ... =
f (a2 ) − f (a1 )
a1 + a2 − 2 · Tep (a)
p e
e
e
e
U
p(m+1)−1 (a) · Up(k+1)−1 (a) − (−1) · Upm−1 (a) · Upk−1 (a)
´
³
´
epk−1 (a)· U
e
e
e
e
e
e
e
U
p(m+1)−1 (a)− Tp (a)· Upm−1 (a) + Upm−1 (a)· Up(k+1)−1 (a)− Tp (a)· Upk−1 (a)
e
U
p(m+k+1)−1 (a)
e
U
p(m+k)−1 (a)
since:
e
=λ
m+k (a) ,
a ∈ R∗ ,
p
e
e
e
e
e
10 ) U
p(m+1)−1 (a) · Up(k+1)−1 (a) − (−1) · Upm−1 (a) · Upk−1 (a) = Up−1 (a) ·
∗
e
U
p(m+k+1)−1 (a), ∀ a ∈ C, ∀ p, m, k ∈ N ;
∗
e
e
e
e
e
20 ) U
p(n+1)−1 (a) − Tp (a) · Upn−1 (a) = Up−1 (a) · Tpn (a), ∀ a ∈ C, ∀ p, n ∈ N ;
∗
epk−1 (a)·Tepm (a)+U
epm−1 (a)·Tepk (a) = U
e
30 ) U
p(m+k)−1 (a), ∀ a ∈ C, ∀ p, m, k ∈ N .
Hence
(6.2)
e
a3 = λ
m+k (a) ,
m, k ∈ N∗ , m 6= k .
e 1 (a) =
An induction argument shows that, if we choose as initial values a1 = λ
(−1)p
∗
e
e
e
(see (2.4)), a ∈ R , the secant method
2 · Tp (a) and a2 = λ2 (a) = 2 · Tp (a) − e
λ1 (a)
gives:
e 1 (a) ,
a1 = λ
e 2 (a) ,
a2 = λ
a ∈ R∗ ,
a ∈ R∗ ,
e
e
e
e
a3 = λ
m+k (a) = λ1+2 (a) = λ3 (a) = λF4 (a) ,
a ∈ R∗ ,
e
e
e
e
a5 = λ
m+k (a) = λ3+5 (a) = λ8 (a) = λF6 (a) ,
a ∈ R∗ ,
e
e
e
a4 = λ
m+k (a) = λ2+3 (a) = λF5 (a) ,
a ∈ R∗ ,
CHEBYSHEV POLYNOMIALS
269
i.e.,
(6.3)
e F (a) −→ λ
e ∞ (a) ,
an = λ
n+1
n
√
√
e ∞ (a) = (a± a2 + 1)p = Tep (a)± a2 + 1·
as the nth approximant to the number λ
ep−1 (a), a ∈ R∗ , where (Fn )n≥0 is the sequence of Fibonacci numbers.
U
In conclusion, we see that for a = 12 and p ∈ N∗ , one obtains the Jamieson’s
results (see (1)) and for a = 21 and p = 1, we derive the Phillips’s results (see
(2)).
REFERENCES
[1] Jamieson, M. – Fibonacci numbers and Aitken sequences revisited, Amer. Math.
Monthly, 97 (1990), 829–831.
[2] Phillips, G. – Aitken sequences and Fibonacci numbers, Amer. Math. Monthly, 91
(1984), 354–357.
[3] Udrea, G. – A problem of Diophantos–Fermat and Chebyshev polynomials of the
second kind, Port. Math., 52 (1995), 301–304.
Gheorghe Udrea,
Str. Unirii-Siret, Bl. 7A, Sc. 1, Ap. 17, Tg-Jiu,
Cod 1400, Judet Gorj – ROMANIA