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How far is 1 + a
nnfrom ea?
Vito Lampret & Philip G Spain
Abstract
We present effective upper and lower bounds for the distance from 1 + a
nnto eafor an
element aof a complex unital Banach algebra and positive integer n. Specifically:
1
2nsup <(z2)e<(z):z∈σ(a).(2)
ea−1 + a
nn
≤kak2
2nekak
where σ(a) is the spectrum of a. The symbol .(p)means “less than or equal to, up to a
term of order n−p”as discussed below.
1 Introduction — technical preliminaries
The purpose of this paper is to establish asymptotic estimates for the quantity
δ(a, n) =
ea−1 + a
nn
(1.1)
where ais an element of a Banach algebra Aand nis a (large) positive integer. We tackle
this problem in three stages: (i) for R[§2], (ii) for C[§3], and (iii) for general A[§4].
The following notation will be useful in presenting our results.
Notation 1. For a fixed positive integer pand functions aand bdefined on Nthe ex-
pression a(n)&(p)b(n)is shorthand for a(n)≥b(n) + O(n−p): that is,
a(n)−b(n)≥M
np(n≥N)
where Mand Nare constants (Npositive) independent of n. The symbols .(p)and '(p)
are defined analogously. These three relations are all transitive.
We shall later extend the use of this notation to instances where the argument nmay run
through the set of half-integers.
Our treatment depends on the estimates of [2, Corollaries 1.1 & 1.2].
0AMS Mathematics Subject Classification: 40A25, 41A99, 46H99, 65D20.
Keywords: Exponential function; Banach algebra; approximation; asymptotic approximation; distance-
estimate; inequalities; hermitian element.
Proposition 1 (Essential Estimates).Consider real numbers xand t. If x > 0and t > x
then
exp x−x2
2t<1 + x
tt<exp x−x2
2(t+x),
while if x < 0and t > |x|we have
exp x−x2
2(t+x)<1 + x
tt<exp x−x2
2t.
Lemma 1. The lengths of the ‘indeterminacy intervals’ of Proposition 1 satisfy
(x > 0) exp x−x2
2(t+x)−exp x−x2
2t
(x < 0) exp x−x2
2t−exp x−x2
2(t+x)
<5
4
|x|3
t2
when t > 2 max{|x|, x2}>0.
Proof. When x > 0 the interval in question has length
exp −x2
2(t+x)−exp −x2
2t
= exp −x2
2texp x2
21
t−1
t+x−1
= exp −x2
2texp x2
2t
x
t+x−1
<exp(r)−1,
where r=|x|3
2t(t+x). Similarly, when x < 0 the interval has length
exp −x2
2t−exp −x2
2(t+x)
= exp −x2
2(t+x)exp x2
21
t+x−1
t−1
= exp −x2
2(t+x)exp x2
2(t+x)
−x
t−1
which again, for the same r,
<exp(r)−1.
Now, under our hypotheses, 2(t+x) = t+ (t+ 2x)≥tand therefore 0 < r < |x|
t
x2
t<1
4.
Hence er−1< r +r2<5
4r < 5
4
|x|3
t2.
vl & pgs 2May 31, 2017
Corollary 1. For any real xwe have
1 + x
mm'(2) exp x−x2
2m.
Specifically: 1 + x
mm−exp x−x2
2m<5
4
|x|3
m2exp(x) (1.2)
when x6= 0 and m > 2 max{|x|, x2}is a (half-)integer.
2 Real case — distance from 1 + x
nnto ex
When xis real we have 1 + x
nn< exfor any positive integer nand therefore (1.1)
simplifies to
δ(x, n) = ex−1 + x
nn.
Lemma 2. For any real xwe have
exp x−x2
2n'(2) ex−x2
2nex.
Proof. It is straightforward to establish that 1 −r+r2
4<exp(−r)<1−r+r2
2for
0< r < 1. Thus, if given x6= 0 and n>x2/2 we define r=x2
2n, then 0 < r < 1 and
exp x−x2
2n−ex−x2
2nex= exp(x)hexp(−r)−1 + ri∈r2
4,r2
2exp(x).
That is x4
16n2ex<exp x−x2
2n−ex−x2
2nex<x4
8n2ex(2.1)
for x6= 0.
Theorem 1. For any real xwe have
δ(x, n)'(2)
x2
2nex.
Specifically, for xreal, 6= 0, and n≥2 max{|x|, x2}we have
δ(x, n)−x2
2nex<|x|3(|x|+ 10)
8n2ex.(2.2)
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Proof. Combining Corollary 1 and Lemma 2 gives
1 + x
nn'(2) exp x−x2
2n'(2) ex−x2
2nex
(recall that '(2) is transitive) and therefore
δ(x, n) = ex−1 + x
nn'(2)
x2
2nex.
More precisely, using (1.2) and (2.1), for x6= 0,
−5
4
|x|3
n2ex<1 + x
nn−exp x−x2
2n<5
4
|x|3
n2ex
and x4
16n2ex<exp x−x2
2n−ex−x2
2nex<x4
8n2ex,
so, adding, we get
x4
16n2ex−5
4
|x|3
n2ex<x2
2nex−δ(x, n)<x4
8n2ex+5
4
|x|3
n2ex=|x|3(|x|+ 10)
8n2ex,
from which (2.2) follows.
Remark 1. This result is more precise than the restriction of Theorem 2 or Theorem 5
to real z.
3 Complex case – distance from 1 + z
nnto ez
Notation 2. For z=x+iy ∈C(that is, x=<(z)and y==(z)) and for n∈Ndefine
m=n
2(so, from now on, mwill be a positive (half-)integer). Write
ξ=x+|z|2
4m∈R&ζ=1 + z
n
2(3.1)
so that
ζ= 1 + x
m+|z|2
4m2= 1 + ξ
m.(3.2)
By the triangle inequality
δ(z, n) = |ez−ζm| ≥ |ez| − 1 + z
n
n
=|ex−ζm|.(3.3)
Further, let
ω(m) = x+y2−x2
4m.(3.4)
vl & pgs 4May 31, 2017
Then
ξ−ξ2
2m=x+y2−x2
4m−x|z|2
4m2−|z|4
32m3=ω(m)−η(m),(3.5)
where
η(m) = x|z|2
4m2+|z|4
32m3=|z|2
32m38mx +|z|2(3.6)
has the same sign as xfor large m(when x6= 0).
Standing assumption
For the rest of this section zwill be a nonzero complex number and ma (half-)integer
such that
m≥max{1,4|z|,4|z|2}.(3.7)
Remark 2. Given z=x+iy ∈Cand a positive half-integer m≥max{1,4|z|,4|z|2}
we have
− |z| ≤ x≤ξ≤ |z|+|z||z|
4m≤17
16 |z|.
Thus
ω≤x+1
16, ξ ≤x+1
16 & 2 max{|ξ|,|ξ|2}< m.
We establish the main result of this section, Theorem 2, by demonstrating the chain of
asymptotic equalities
ζm'(2) exp ξ−ξ2
2m'(2) exp(ω(m)) '(2) ex1 + y2−x2
4m.
Lemma 3.
ζm'(2) exp ξ−ξ2
2m.
Proof. Apply (1.2) (with ξin place of x). Then, bearing in mind Remark 2, and also the
arithmetical fact that 5
417
16 3e1
16 <1.6, we have
ζm−exp ξ−ξ2
2m<5
4
|ξ|3
m2exp(ξ)
<5
417
163|z|3
m2exp x+1
16<2|z|3
m2exp (x).
Lemma 4.
exp ξ−ξ2
2m'(2) exp(ω(m)).
Proof. According to (3.6), we have
|η| ≤ |z|3
4m2+|z|4
32m3=|z|3
4m21 + |z|
8m
≤|z|3
4m21 + 1
2=3
8
|z|3
m2≤3
128 <1,(3.8)
vl & pgs 5May 31, 2017
from which |e−η−1|<2|η|. Thus, using (3.5), (3.4), and (3.8), together with Remark 2,
exp ξ−ξ2
2m−exp (ω)
(3.5)
≤exp (ω)e−η−1
(3.4)
≤e−η−1exp 1
16 exp (x)
≤2|η|exp 1
16 exp (x)(3.8)
≤2 3
8
|z|3
m2!exp 1
16 exp(x)
≤|z|3
m2exp(x).
Lemma 5.
exp(ω(m)) '(2) ex1 + y2−x2
4m.
Proof. Recall that |er−1−r| ≤ r2(−1< r < 1).With r=y2−x2
4mwe have |r| ≤
|z|2
4m≤1
16, and therefore
exp y2−x2
4m−1−y2−x2
4m≤|z|4
16m2,
from which exp x+y2−x2
4m−ex1 + y2−x2
4m≤|z|4
16m2ex.
Theorem 2. Given z=x+iy ∈Cwe have
δ(z, n) = ez−1 + z
nn&(2)
|<(z2)|
2ne<(z).
Specifically:
δ(z, n)≥|<(z2)|
2ne<(z)−12 |z|3
n21 + |z|
48 e<(z)
for n≥2 max{1,4|z|,4|z|2}.
Proof. Since m > 2 max{|ξ|,|ξ|2}, see Remark 2, we can apply Lemmas 3, 4 and 5 to
get
ζm−ex−y2−x2
4mex≤ζm−exp ξ−ξ2
2m
+exp ξ−ξ2
2m−exp (ω)
+exexp y2−x2
4m−1−y2−x2
4m
≤2|z|3
m2exp(x) + |z|3
m2exp(x) + |z|4
16m2exp(x)
=M
m2
vl & pgs 6May 31, 2017
where M= 3 |z|31 + |z|
48 exdepends only on z. Now
|ζm−ex| − |y2−x2|
4mex≤ζm−ex−y2−x2
4mex≤M
m2
and therefore
δ(z, n)(3.3)
≥ |ζm−ex| ≥ |y2−x2|
4mex−M
m2.
Remark 3. The upper bound
δ(z, n) = ez−1 + z
nn≤|z|2
2ne|z|,
valid for z∈Cand n > |z|, is the scalar variant of Theorem 5 below. Note the presence
of the term e|z|in contrast to the exof Theorem 1.
When zlies on an (anti)diagonal
Theorem 2 tells us little when <(z2) = 0, that is, when zlies on X, the union of the
diagonal and antidiagonal in C:
X={z∈C:x2=y2}={z∈C:<(z2) = 0}.
We can, however, then derive a higher order estimate for the lower bound.
Theorem 3. Let z=x+iy (6= 0) ∈Cwith x2=y2. Then
δ(z, n)&(3)
|<(z)|3
2n2e<(z).
Specifically,
ez−1 + z
nn>|x|3
2n2ex−|x|x3(1 + x)
n3ex(n > 4|x|).
Proof. Write l=x
m. Then, by (3.7) and (3.2), we have
|l|<1 & ζ= 1 + ξ
m= 1 + l+l2
2.(3.9)
Let Ibe the interval with end-points eland ζ. Then
δ(z, n)≥(el)m−ζm
≥el−ζmin
r∈Id
dr(rm)
=mel−ζmin exp m−1
mx, ζm−1.
vl & pgs 7May 31, 2017
Now
el−ζ(3.9)
=
l3
3! +l4
4! +l5
5! +. . . =
l3
3! 1 + l
4+l5
5! 1 + l
6+. . .
(3.9)
=|l|3
3! 1 + l
4+|l|5
5! 1 + l
6+. . . ≥3
4
|l|3
3! =|l|3
8=|x|3
8m3.
Further, recalling the definitions (3.1) and (3.2), we see that ξ > x. Moreover, 0 <
1 + x
m<1 + ξ
m=ζfor m > |x|. Thus, by Proposition 1,
ζm−1>1 + x
mm−1
m
>exp m−1
mx−x2
2m
>exp x−x
m−x2
2mif m > x > 0,
while
ζm−1>exp m−1
mx−x2
2(m+x)
>exp x−x
m−x2
mif m > −2x > 0.
Hence
min exp m−1
mx, ζm−1>exp x−x
m−x2
m
and therefore
δ(z, n)>|x|3
8m2exp x−x
m−x2
m(m > 2|x|)
≥|x|3
8m2ex1−x(1 + x)
m.
Supremum of {δ(z, n) : z∈K}
Note that
δ(z, n)≤ |ez|+1 + ξ
mm
≤ex+eξ=ex1 + e|z|2
4m≤ex1 + e|z|,
showing that the following is a good definition:
Definition 1. Given a bounded subset Kof Cdefine
∆(K, n) = sup{δ(z, n) : z∈K}
for each n∈N.
vl & pgs 8May 31, 2017
Combining the results of Theorems 2 and 3 yields
Theorem 4. Let Kbe a bounded subset of C. Then, asymptotically, to within a term of
the order n−2, and uniformly over K,
∆(K, n)&(2)
1
2nsup <(z2)e<(z):z∈K.
If K⊂ X then, asymptotically, to within a term of order n−3, and uniformly over K,
∆(K, n)&(3)
1
2n2sup |<(z)|3e<(x):z∈K.
4 How far is 1 + a
nnfrom eain a Banach algebra?
Upper bound for δ(a, n)
Theorem 5. Let (A,k·k)be a real or complex norm-unital Banach algebra and a∈ A
with a6= 0. Then, for every integer n > kak,
δ(a, n) =
ea−1 + a
nn
≤ekak−1 + kak
nn
< ekak"1−exp −kak2
2n!#
<kak2
2nekak.
Proof. Following [1, §8], for every n∈Nwe define the partial sums and powers
sn=
n
X
k=0
ak
k!, σn=
n
X
k=0
kakk
k!, bn=1 + a
nn, βn=1 + kak
nn
.(4.1)
Then
lim
n→∞sn= exp(a) = lim
n→∞bn& lim
n→∞σn= exp(kak) = lim
n→∞βn.(4.2)
Using the binomial expansions for bnand βn, and introducing the notation
λk(n) = 1
k!1−n!
nk(n−k)!
for every integer ksuch that 0 ≤k≤n, we have
sn−bn=
n
X
k=0
1
k!1−n!
nk
1
(n−k)!
| {z }
=λk(n)≥0
ak=
n
X
k=0
λk(n)ak(4.3)
and
σn−βn=
n
X
k=0
λk(n)kakk(4.4)
vl & pgs 9May 31, 2017
with obviously all λk(n)≥0, where
λ0(n)≡λ1(n)≡0 (n≥1) and λ2(n)≡1
2n(n≥2) .
Using the triangle inequality we estimate
kea−bnk=k(ea−sn)+(sn−bn)k≤kea−snk+ksn−bnk.
Now, considering (4.1)–(4.2),
kea−snk=
∞
X
k=n+1
ak
k!
≤
∞
X
k=n+1
kakk
k!=ekak−σn,
and, referring to (4.3) and (4.4),
ksn−bnk=
n
X
k=0
λkak
≤
n
X
k=0
λkkakk=σn−βn.
Consequently
kea−bnk ≤ ekak−σn+ (σn−βn) = ekak−βn.
Therefore, from (4.1), we obtain the estimate
ea−1 + a
nn
≤ekak−1 + kak
nn
which, see Proposition 1,
< ekak−ekakexp −kak2
2n!
and this
<kak2
2nekak
since 1 −e−r< r (r∈R).
Asymptotic lower bound for δ(a, n)
We write, as usual, σ(a) for the spectrum and |a|σfor the spectral radius of an element a
of a complex norm-unital Banach algebra.
Theorem 6. Let Abe a norm-unital complex Banach algebra. Then
δ(a, n) =
ea−1 + a
nn
≥ea−1 + a
nnσ
≥∆(σ(a), n)
&(2)
1
2nsup <(z2)e<(z):z∈σ(a)
for any a∈ A and positive integer n.
Thus, if ais not quasinilpotent,
ea−1 + a
nn
tends to 0no faster than O(n−1)unless
σ(a)is contained in Xin which case the rate of convergence is no faster than O(n−2).
vl & pgs 10 May 31, 2017
Proof. Without loss of generality we may assume that Ais commutative. Then
σ(a) = {φ(a) : φ∈Φ(A)}
where Φ(A) is the set of characters (nontrivial multiplicative functionals) on A.
Now, for any φ∈Φ(A) and any n(recall that the norm dominates the spectral radius),
eφ(a)−1 + φ(a)
nn=φea−1 + a
nn
≤ea−1 + a
nnσ
≤
ea−1 + a
nn
.
The rest follows from Theorem 4.
Remark 4. The first inequality in Theorem 5 is sharp: we have equality if ais a positive
real in C.
Theorem 6 too is sharp: if a6= 0 but a2= 0 then ea=1 + a
nnfor n≥1.
Hermitian elements of a Banach algebra
We refer to [1, §10] for the background on numerical range in Banach algebras. Recall
that an element hof a complex norm-unital Banach algebra is hermitian if its algebra
numerical range is real: equivalently, if
eirh
= 1 (r∈R): equivalently, if k1 + irhk ≤
1 + o(r) (R3r→0). In analogy to Theorem 1, combining Theorems 5 & 6:
Theorem 7. Let hbe a hermitian element of a complex unital Banach algebra. Then
khk2
2ne−khk.(2)
eh−1 + h
nn
≤khk2
2nekhk.
When khk ∈ σ(h), then, more precisely,
khk2
2nekhk.(2)
eh−1 + h
nn
≤khk2
2nekhk.
Proof. For such an hwe have σ(h)⊂Rand khk=|h|σ: so then at least one of khkand
− khkbelongs to σ(h). Thus, referring to Theorem 6,
sup <(z2)e<(z):z∈σ(a)≥ khk2e−khkif − khk ∈ σ(h)
and
sup <(z2)e<(z):z∈σ(a)≥ khk2ekhkif khk ∈ σ(h).
Remark 5. Our upper estimate (Theorem 5) for the size of ea−1 + a
nnholds for
an element of a real or complex norm-unital Banach algebra, while for the lower bound
(Theorem 6) we require the algebra to be complex. Qualitatively speaking, this is no
vl & pgs 11 May 31, 2017
essential restriction because: (i) if a Banach algebra Ahas a unit 1whose norm is 6= 1
we can construct an equivalent Banach algebra norm on Afor which k1k= 1; and (ii) if
Ahas no unit we can adjoin a unit and renorm so that this unit has norm 1; and (iii)
if Ais a Banach algebra over the real field we can embed it isometrically in a complex
Banach algebra (see [1, §§3, 4 & 13]). Note that the expression ea−1 + a
nncan be
interpreted as the limit of the sequence
N
X
k=1 ak
k!−n
kak
nkas N→ ∞ in any Banach
algebra (even if it is neither unital nor complex).
References
[1] F. F. Bonsall & J. Duncan, Complete Normed Algebras, Springer (1973).
[2] V. Lampret, Approximating the powers with large exponents and bases close to unit,
and the associated sequence of nested limits, Int. J. Contemp. Math. Sci. 6(2011)
2135–2145.
Vito Lampret, University of Ljubljana, Ljubljana, Slovenia 386 EU
vito.lampret@guest.arnes.si
&
Philip G Spain, Department of Mathematics and Statistics, University of Glasgow, Glas-
gow G12 8QW
Philip.Spain@glasgow.ac.uk
vl & pgs 12 May 31, 2017