Preliminaries: two functions we will compose, an a1 t0 (u ) = u + b0 ; tn (u ) = ; t0 t1 t 2 (0) = b0 + ; φ n = C (n ) = (t0 t1 ... t n )(0) a bn + u b1 + b2 +2 0
(1.1)
In Mathematica, the (1.1) is realized as, an
t[u _, n _] :=
bn + u
a3
[0, 0 ], 1], 2], 3] = → t[t [t[t [0
a2
+ b3
a1 a0 b0
for tn (u ) =
an bn +u
=
d n an dnbn +d nu
+ b2
+ b1
, you have,
φ n = b0 +
a1 b1 +
=ɺ b0 +
a2 a
b2 +⋱ bn
a1
a2
b1 + b2 +
. ..
an
= b0 + K nn′=1 (
bn +
an′ bn′
)=
A( n) B ( n)
n
n′ = n
= C (n) = φ n = C ( n) = ( t0 t1 ... tn )(0) = b0 −
∑
∏
n′′ = n′ n′′ =1
(1.2) (−an′′ )
Bn′−1 Bn′
n′=1
We will also need theorem 9.2, 1
th Theorem 9.2 : Consider the continued fraction C = b0 + K ( bnn ) with n approximant C(n)=A(n)/B(n). Then A(n) a
and B(n) satisfy, respectively, the difference equations: A( n) = bn A(n − 1) + an A(n − 2); A( −1) = 1; A(0) = b0 ; B ( n) = bn B (n − 1) + an B (n − 2); B ( −1) = 0; B (0) = 1; → C ( n) =
A( n) B( n)
bn A( n − 1) + an A( n − 2)
=
bn B( n − 1) + an B (n − 2)
n
= b0 + K n′=1 (
an′ bn′
(1.3) )
Lastly: we will need: Remark 9.3c: if K (a∞ , −b∞ ) converges to x(0) / x( −1) , then K (a∞ , +b∞ ) converges to − x(0) / x( −1) , which is to say, K(
a∞ −b∞
)=
x (0) x ( −1)
→ K (
a
a∞ +b∞
)=−
x ( 0)
(1.4)
x ( −1) a′
a
a′
Equivalency, “ ≈ ”: Two continued fractions K ( bnn ) = b0 + K nn′=1 ( bnn′′ ) and K ( bn′n ) = b0′ + K nn′=1 ( bn′n′′ ) are said to be a
a′
“equivalent” if they have the same sequence of approximants: K ( bnn ) ≈ K ( bnn′ ) . ∃{d n′ }n∞′=1 → tn (u ) =
an bn + u
=
dn an d n bn + d nu
= sn rn ; sn (u ) =
d n an d nbn + u
; rn (u ) = d nu ;
The function-composition-operation associates; therefore, we have, t1 t2 ... tn = ( s1 r1 ) ( s2 r2 ) ... ( sn rn ) = s1 (r1 s2 ) r2 ... (rn −1 sn ) rn ;
(1.5)
Define tn* = t n* (u ) ,
1
Converse of Th. 9.2 is true: every homogeneous second-order difference equation gives rise to an associated continued fraction!
(1.6)
tn* (u ) ≡ rn −1 sn =
d n −1dn an d n bn + u
→ C (n) ≡ (t1 t2 ... tn )(0) = (t1* t 2* ... tn* )(0)
(1.7)
Equivalence relation, C (n) = C (n* ) → K (
an bn
) ≈ K(
d n −1d n an ) dn bn
(1.8)
Example: consider ∃{d n′ }n∞′=1 , and d n bn = 1 . Then, (1.8) says K (an / bn ) ≈ K (bn −1bn an / 1) . Example: consider ∃{d n′ }n∞′=1 , and d 1 = d 2n =
1 a1
1 a2 n d 2 n −1
, d n = =
1 an d n−1
. Then, (1.8) says K (an / bn ) ≈ K (1/ bn dn ) .
a1a3 ...a2 n−1 a2 a4 ...a2 n
1
; d 2n +1 =
a2 n +1 d2 n
=
a2 a4 ...a2 n a1a3 ...a2n +1
(1.9)
;
NOTE: the steps that lead from (1.8) to (1.9) are left to: 406 - pr 08 - proof of even and odd terms in dn mobius sequence. Theorem 9.4: Let bn > 0 , n ∈ ℤ + . Then the continued fraction K (1/ bn ) is convergent if and only if the infinite series
∑
∞
th
b is divergent . Proof: using the closed-form expression for the n term in a continued-fraction-
n =1 n
sequence, indicated in (1.2), we compute K (1/ bn ) as, K (
1 bn
n′ = n
)=−
∑
∏
n′ =1
n′′=n′ n′′=1
(−1)
Bn′ −1Bn′
→ Bn +1 > Bn −1 → Bn Bn +1 > Bn −1 Bn
alternating Bn = Bn −2 + bn ⋅ Bn −1 ; =∑ = ; B(0) = 1; B(1) = b ; → series n′=1 Bn′ −1 Bn ′ 1 n ′ =n
′
(−1)n +1
Because the summand’s magnitude
( −1)n+1 Bn −1Bn
n +1
( −1) Bn−1 Bn
monotonically-decreases
(1.10)
monotonically decreases, the corresponding series converges iff .,
lim n →∞ Bn−11 Bn = 0 . Note this condition can be re-expressed as, 2 2 Bn−1Bn = Bn − 2 Bn −1 + bn Bn −1 ≥ Bn−2 Bn −1 + bn γ ≥
(∑
n′ = n
)
2 b γ → n′ =1 n ′
∑
n′ = n n′ =1
bn ′ = ∞ ↔ limn →∞ Bn−11Bn = 0
(1.11)
Note: from the recursion Bn = Bn − 2 + bn ⋅ Bn −1 of (1.10) with the indicated initial-conditions, you have, n′= n
induction → Bn− 1 + Bn ≤ ∏ n′ (1 + bn′ ) < exp( Bn −1 + Bn = Bn− 2 + (1 + bn ) Bn−1 ≤ (1 + bn ) ⋅ ( Bn−1 + Bn− 2 ) =1
Suppose:
∑
n ′=∞ n ′=1
L
2
L 2
bn′ = L → Bn−1 + Bn < e → Bn−1 + Bn ≤ ( Bn−1 + Bn ) ≤ (e ) = e 1 4
1 4
1 4
2L
∑
n ′= n n ′ =1
bn′ );
→ lim Bn−1Bn ≠ 0 → divergence 1
(1.12)
n →∞
2
Theorem 9.5 : consider, again, the following ubiquitous and general difference equation, xn − bn xn −1 − an xn− 2 = 0; an ≠ 0; The
a1 a2 a3 b1 + b2 + b3 +
(1.13)
A ( n ) ... = lim B ( n ) − b0 continued fraction converges iff the abovementioned difference equation has a n →∞
minimal solution ϕ (n) with ϕ (0) ≠ 0 . In case of convergence, moreoever, one has: ϕ
− ϕ n−1 = n− 2
an
an+1
a n+ 2
bn + bn+1 + bn+2 +
... =
A∞ B∞
A
− Bn−1 − b0
(1.14)
n −1
A ( n )
Proof: assume the continued fraction lim B ( n ) − b0 = C ( ∞) − b0 converges. Hence: if A(n) and B(n) are the n
th
n →∞
th
A ( n )
partial numerator and n partial denominator of this C (∞) − b0 , respectively, then: C (∞ ) = C∞ = lim B ( n ) ≡ L . n →∞
2
Also known as: “Pincherle’s theorem”.
Then: (1.3) says A( n) = An and B( n) = Bn satisfy the recursion relation (1.13) if we use the initial conditions A(−1) = 1 , A(0) = 0 , and B(−1) = 0 , B (0) = 1 . Claim of minimal solution: Fact: ϕ (n) = A( n) − LB( n) is a minimal solution of x − b x n
n
n
−1
−a x n
n
−2
= 0 . Proof:
let y ( n) = α ⋅ A( n) + β ⋅ B( n) for scalars α , β , and let y( n) be some other solution of (1.13). then, lim
n →∞
ϕ (n) y (n)
= lim
n →∞
A(n) − LB(n)
α A(n) + β B(n)
= lim
n →∞
A( n ) B ( n )
α
− L
A ( n ) B ( n )
+β
≡
L−L
αL + β
= 0; ϕ (−1) = 1 ≠ 0;
(1.15)
Claim of minimal solution: Fact: the (1.13) has a minimal solution ϕ ( n) with ϕ ( −1) ≠ 0 . From the
convergence-limit
∑
n′= n n′=1
3
bn′ = ∞ ↔ lim n →∞ Bn 11Bn = 0 of (1.11), the “associated continued fraction” with the −
difference-equation (1.13) is Cn − b0 = K (
an − bn
*
th
A ( n ) * ) with the n approximant C * ( n) = B ; in which (n) and A * (n)
B* (n) are two linearly-independent solutions to (1.13) (c.f., (1.3)) if we use initial conditions A* (−1) = 1 , A* (0) = 0 , and B* (−1) = 0 , B* (0) = 1 . Consequently: ϕ (n) = A* (n) − LB * (n) for n ≥ 0 . We then observe that,
0 = lim
n →∞
ϕ ( n) *
B ( n)
A* ( n) − LB* ( n)
= lim
*
B ( n)
n →∞
= lim n →∞
A* ( n) *
B ( n)
− L → lim
n →∞
A* ( n) *
B ( n)
= L=
ϕ (0) ϕ ( −1)
(1.16)
Using (1.4), we get that, A∞ B∞
= C∞ = lim C (n) = lim n →∞
n→∞
A( n) B ( n)
= −
ϕ (0) ϕ ( −1)
(1.17)
This proves the first part of the theorem. It also proves (1.14) for n = 1 . The proof of (1.14) for all n is found in 405 - pr 05 - proof of (9-2-10) for all n. Example: consider the continued fraction C ∞ = 1a+ 1a+ 1a+ ... , where a ∈ ℂ . Find conditions on a under which the continued-fraction converges. There are actually several ways to do this. th
Method 1: Use (1.3) of Theorem 9.2; to that end, let A(n) and B(n) be the n partial numerator and denominator, respectively, and, A(−1) = 1; A(0) = b0 = 0; an = a; A( n) = bn A( n − 1) + an A( n − 2) = A( n − 1) + aA( n − 2); → B (−1) = 1 ≠ 0; B(0) = 1; bn = 1; B( n) = bn B( n − 1) + an B( n − 2) = B( n − 1) + aB( n − 2);
(1.18)
Using the ansatz A( n) = λ n , we get the quadratic, n
A(n) = λ = B( n) →
λ n = λ n−1 + aλ n −2 n
λ =λ
n−1
2 1 → λ − λ − a = 0 → λ± = 2 (1 ± 1 + 4 a ) = λ1,2 + aλ n −2
(1.19)
Break up into the following cases: consider the case of (1) λ1 ≠ λ 2 vs. (2) λ1 = λ 2 , where λ2 < λ 1 . For (1), we have (1.13) appear as xn − xn −1 − axn −2 = 0 , and we see it has a “minimal” solution. Consequently: by Theorem 9.5 realized as (1.16), our continued fraction C ∞ = 1a+ 1a+ 1a+ ... converges to −λ − = − 12 (1 − 1 + 4a ) . For (2), you have degenerate roots if a =
3
See… um, I forget.
−1 4
→ λ1 = λ 2 =
1 2
n
n
→ An = c1 ( 12 ) + c2 n( 12 ) . Using the initial conditions
already-indicated in (1.18), A( −1) = 1 and A(0) = b0 = 0 , we have (c1 , c2 ) = (0, −21 ) , and so An = − n( 12 )n+1 . We could use the B-equation of (1.18) and get: Bn = (n + 1)( 12 ) n ; hence, a A a 1 + 1 + 4a −1 K ( ) = lim n = = − λ 2 ; a ∈ ℂ → K ( ) = − ↔ a∉ {x∈ ℝ : x < n →∞ B 1 2 1 2 n
}
(1.20)
≤ a ∈ ℝ → x ∈ ℝ;
(1.21)
−1 4
Method 2 (less formal): just assume the limit of the series exists, and you have, a a a
1+ 1+ 1+
... = x =
a
1 + x
2
→ x + x − a = 0 → x± =
1 ± 1 + 4a 2
;
−1 4
The problem with this 2 nd method is that it assumes what we are trying to prove.
