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two steps: First we apply the Splitting Lemma, decoupling the system
into two one-dimensional systems. Then we apply Theorem 11 (p. 52)
to both of these systems. From Exercise 1 on p. 55 we conclude that
we are able to take N = 1 and M = rв€’1 = 0. This shows that there is
an HLFFS for which the transformed system has a coeп¬ѓcient matrix
Solutions 251

of the form B(z) = zA0 + diag [a, d], and then F (z) = I + 1 Fn z в€’n
Л†
can be computed directly as follows: Denoting the entries in the п¬Ѓrst
column of Fn by fn and gn , they satisfy the recursions
n fn = в€’b gn , (О»1 в€’ О»2 ) gn+1 = (n + d в€’ a) gn + c fn , n в‰Ґ 1,
with g1 = c/(О»1 в€’ О»2 ). Setting О± + ОІ = d в€’ a, О± ОІ = в€’b c, we obtain
пЈј
fn = (О»1 в€’О»(ОІ)nn!
(О±)n
пЈЅ
n
2)
, n в‰Ґ 1.
(1+О±)nв€’1 (1+ОІ)nв€’1 пЈѕ
gn = c (О»1 в€’О»2 )n (nв€’1)!

Similarly, one can obtain the other two entries of Fn .
In case (b), use a scalar exponential shift to eliminate О», then use a
constant transformation T that commutes with A0 and produces a
new matrix A1 for which the (new) element c vanishes. Then, use the
transformation
b1/2 в€’b1/2
1
x = diag [1, z /2] x
Лњ
1 1
and the change of variable z = w2 to obtain a system of the form
treated in case (a).

Section 4.1:
1. Note that by deп¬Ѓnition of sectorial regions it suп¬ѓces to prove the
statement for sectors.
2. Use the previous exercise to see that it suп¬ѓces to deal with k = 1 and
П„ = 0, in which case the inequalities describe a circle with center on
the positive real axis, touching the origin, resp. the right half-plane
in case of c = 0.
3. Let I = (a, b); then the union has opening О± + b в€’ a and bisecting
direction (b + a)/2.

Section 4.2:
z
1. Use [f (z) в€’ f (0)] z в€’1 в€’ f (0) = z в€’1 (u) в€’ f (0)] du.
[f
0

2. Conclude that diп¬Ђerentiability at the origin is equivalent to f (z) =
f (0)+z f (0)+z h(z), with h(z) continuous at the origin and h(0) = 0.
Then, use CauchyвЂ™s integral formula for f (z).
3. Use the fact that, according to the previous exercises, diп¬Ђerentiability
at the origin is equivalent to continuity of the derivative at the origin.
4. Justify termwise integration of the power series expansion of ezt .
252 Solutions

Section 4.3:

6. Proceed as in Exercise 2 on p. 41 to show that E [[z]]s is closed with
respect to multiplication. Moreover, use (B.13) to derive that E [[z]]s
is closed with respect to derivation.

7. Compare Exercise 3 on p. 41.

8. Use (B.13) (p. 231).

Section 4.4:
z в€’w
1. Observe 1 в€’ eв€’z = e dw and estimate the integral.
0

5. Use the previous two exercises.

6. Use Proposition 8 (p. 66).

Section 4.5:

2. Use Exercise 1.

3. Use (B.13) (p. 231).
N в€’1 N в€’mв€’1
5. Observe z m f (z) в€’ fnв€’m z n = z m (f (z) в€’ fn z n ) and
n=0 n=в€’m
use (B.13) (p. 231).

6. To show that every element of As,m (G, C ) is invertible, observe that
the number of zeros of f (z) in an arbitrary closed subsector of G is
п¬Ѓnite.
Л†
7. Take any f в€€ E [[z]]s в€Ђs в‰Ґ 0, and observe RittвЂ™s theorem.

Section 4.6:

2. For (a), use the same arguments as in the proof of Proposition 7
(p. 65) to show gn в‰¤ cn for every n в‰Ґ 0. For (b), use the integral
representation for the Gamma function to show
в€ћ(d)
k
z N rf (z, N ) = z в€’k uN rg (u, N ) eв€’(u/z) duk
0

and estimate as usual.

4. For x в‰Ґ 0, п¬Ѓnd the maximum of xn eв€’x and use StirlingвЂ™s formula.
Solutions 253

6. Abbreviate s = 1/k. By assumption there exist c, K > 0 so that
ВЇ
f (z) в‰¤ c (|z|K)n О“(1 + sn) for every n в‰Ґ 0 and z в€€ S. According to
StirlingвЂ™s formula the same holds, for diп¬Ђerent constants c, K, with
О“(1 + sn) replaced by nsn . For small |z|, the term (|z|K)n nsn п¬Ѓrst
decreases, then increases with respect to n and is approximately min-
imal for n with |z|K es ns в‰¤ 1 < |z|K es (n + 1)s . For this n we have
Лњ в€’k
(|z|K)n nsn в‰¤ eв€’sn в‰¤ ceв€’K|z| , with suп¬ѓciently large constants
Лњ
ЛњЛњ
c, K, independent of z.

7. Use the previous exercise.

Section 4.7:

2. Let G = {z в€€ G : ze2ПЂi в€€ G}. Conclude that f (z) в€’ f (ze2ПЂi ) в€јs Л† in
Лњ =0
Лњ and then use WatsonвЂ™s Lemma and Proposition 7 (p. 65).
G

Section 5.1:

1. Make a change of variable (u/z)k = x in (5.1). The formula holds
whenever the integral exists, which is certainly the case for О» = 0.

2. Make a change of variable (u/z)k = x in (5.1) and estimate f .

3. Substitute uk = ak + tk , conclude that f (t) = f ((ak + tk )1/k ) is
Лњ
of exponential growth at most k on the line arg t = П„ , and use the
previous exercise.

Section 5.2:

1. Use (B.10) (p. 228).

2. Estimate the integral for g = Bk f in terms of the length of the path
of integration times maximum of the integrand, occurring along the
circular part of Оі(П„ ), to obtain g(u) в‰¤ c exp[KПЃОє1 + (|u|/ПЃ)k ], with
ПЃ being the radius of the circular part. Then, observe that ПЃ can be
selected, depending upon |u|, so that the right-hand side is minimal.

3. Make a change of variable z в€’k = w, both in (5.4) and (5.3).

Section 5.3:

2. For real t > 0, deп¬Ѓne
в€ћ(В±П„ )
du
В±
f (u) eв€’uz
h (z) = ,
uв€’t
0
254 Solutions

with suп¬ѓciently small П„ > 0. Use Theorem 22 (p. 79) to show that
hВ± (z) is holomorphic in a region GВ± containing the straight line
z(y) = x0 + iy, with suп¬ѓciently large x0 (п¬Ѓxed) and в€’в€ћ < y <
в€ћ, and h+ (z(y)) (resp. hв€’ (z(y)) tends to zero as y в†’ в€’в€ћ (resp.
y в†’ +в€ћ). For z(y) as above, check that hВ± (z(y)) can be represented
by an integral as above, but integration along the positive real axis,
with a small detour about t, say, along a half-circle, in the upper,
resp. lower, half-plane. Using CauchyвЂ™s integral formula, show
hв€’ (z(y)) = h+ (z(y)) + 2ПЂif (t) eв€’tz(y) .
From this вЂњconnection relation,вЂќ conclude
lim hВ± (z(y))etz(y) = в€“2ПЂi f (t).
yв†’В±в€ћ

Then, observe that the left-hand side of (5.5) can be seen to equal
(2ПЂi)в€’1 [h+ (z(в€’y))в€’h+ (z(y))]. From these identities, conclude (5.5).

Section 5.4:
N в€’1
1. Observe n=1 (ws /z)n = (wsN z 1в€’N в€’ws ) (ws в€’z)в€’1 , and use (B.10)
(p. 228) and (B.14) (p. 232).
2. Observe (B.10) (p. 228). Can you show that the asymptotic expansion
is of Gevrey order s by estimating the integral representation for
Es (z; N )?

Section 5.5:
1. Compute the moment function corresponding to e(z; О±) and deп¬Ѓne
E(z; О±) analogous to (5.12). Then, derive the integral representation
1 dw
wв€’О± E(w)
E(z; О±) = ,
wв€’z
2ПЂi Оі

for |z| < 1, and discuss the behavior for z в†’ в€ћ.
3. Analyze the behavior of E(z; О±) as z в†’ в€ћ, following the same argu-
ments used in the proof of Lemma 6 (p. 84), and then proceed as in
the proof of Theorem 26 (p. 84).

Section 5.6:
2. Show the integral representation
в€ћ(П„ )
du
k(z) = e1 (u/z) E2 (u) ,
u
0

with 2k|П„ | < ПЂ, and discuss convergence of the integral, using the
behavior of E2 (u) as u tends to inп¬Ѓnity.
Solutions 255

3. Use the fact that the moment functions mj (u/p) correspond to the
kernels p ej (z p ), which are of order kp. Moreover, show
pв€’1
kp ([z e2jПЂi ]1/p ),
p k(z) = 0 < arg z < 2ПЂ.
j=0

Section 5.8:
3. Use Lemma 7 (p. 94).
4. StirlingвЂ™s formula (B.12) (p. 229) implies that the radius of conver-
gence equals 1. Holomorphic continuation follows from Exercise 3 on
p. 90.

Section 6.1:
ОЅ m
1. Show the functional equations g(z) = ОЅ=0 z 2 + g(z 2 ), m в‰Ґ 1.
mв€’1

Use this to show that g(r) cannot have a limit as r в†’ 1в€’. Then,
conclude that g(z) cannot have a radial limit at any one of the points
zjm = exp[2jПЂi2в€’m ], for arbitrary j в€€ N.
2. If E contains at least one a = 0, multiply all the terms of the power
Л†
series f (z) by a.

Section 6.2:
1. Observe (S (Bk f ))(u) = (1 в€’ u)в€’1 .
Л†Л†

5. Use Exercises 4вЂ“7, p. 63.
в€ћ
6. For part (a), show (1 в€’ 4z)в€’1/2 = О“(1 + 2n) [О“(1 + n)]в€’2 z n , in
n=0
the disc |z| < 1/4.

Section 6.3:
1. Show that (e в€’ a)n is a Cauchy sequence, and hence deп¬Ѓnes some
b в€€ E . Then, verify a b = b a = e.
2. Show п¬Ѓrst that it suп¬ѓces to consider the case of f0 = e. Then, con-
ВЇ
clude for every closed subsector S of G with suп¬ѓciently small radius
ВЇ
that e в€’ f (z) < 1 for every z в€€ S, and apply the previous exercise.
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