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## Abstract

We present effective upper and lower bounds for the distance from $\displaystyle \left(1 + \frac{a}{n}\right)^n$ to $\displaystyle e^a$ for an element $a$ of a complex unital Banach algebra and positive integer $n$. Specifically: $\frac1{2n} \sup \left\{ \abs{\Re (z^2) } e^{\Re (z)} : z \in \sigma(a) \right\} \ \lesssim_{(2)} \ \nm{e^a - \left(1 + \frac{a}{n}\right)^n} \ \le \ \frac{\nm{a}^2}{2n} \ e^{\nm{a}} .$
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):zσ(a).(2)
ea1 + 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 npas discussed below.
1 Introduction — technical preliminaries
The purpose of this paper is to establish asymptotic estimates for the quantity
δ(a, n) =
ea1 + 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(np): that is,
a(n)b(n)M
np(nN)
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 xx2
2t<1 + x
tt<exp xx2
2(t+x),
while if x < 0and t > |x|we have
exp xx2
2(t+x)<1 + x
tt<exp xx2
2t.
Lemma 1. The lengths of the ‘indeterminacy intervals’ of Proposition 1 satisfy
(x > 0) exp xx2
2(t+x)exp xx2
2t
(x < 0) exp xx2
2texp xx2
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
t1
t+x1
= exp x2
2texp x2
2t
x
t+x1
<exp(r)1,
where r=|x|3
2t(t+x). Similarly, when x < 0 the interval has length
exp x2
2texp x2
2(t+x)
= exp x2
2(t+x)exp x2
21
t+x1
t1
= exp x2
2(t+x)exp x2
2(t+x)
x
t1
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 er1< r +r2<5
4r < 5
4
|x|3
t2.
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Corollary 1. For any real xwe have
1 + x
mm'(2) exp xx2
2m.
Speciﬁcally: 1 + x
mmexp xx2
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) = ex1 + x
nn.
Lemma 2. For any real xwe have
exp xx2
2n'(2) exx2
2nex.
Proof. It is straightforward to establish that 1 r+r2
4<exp(r)<1r+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 xx2
2nexx2
2nex= exp(x)hexp(r)1 + rir2
4,r2
2exp(x).
That is x4
16n2ex<exp xx2
2nexx2
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 n2 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 xx2
2n'(2) exx2
2nex
(recall that '(2) is transitive) and therefore
δ(x, n) = ex1 + x
nn'(2)
x2
2nex.
More precisely, using (1.2) and (2.1), for x6= 0,
5
4
|x|3
n2ex<1 + x
nnexp xx2
2n<5
4
|x|3
n2ex
and x4
16n2ex<exp xx2
2nexx2
2nex<x4
8n2ex,
x4
16n2ex5
4
|x|3
n2ex<x2
2nexδ(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 nNdeﬁne
m=n
2(so, from now on, mwill be a positive (half-)integer). Write
ξ=x+|z|2
4mR&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) = |ezζm| ≥ |ez| − 1 + z
n
n
=|exζm|.(3.3)
Further, let
ω(m) = x+y2x2
4m.(3.4)
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Then
ξξ2
2m=x+y2x2
4mx|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
mmax{1,4|z|,4|z|2}.(3.7)
Remark 2. Given z=x+iy Cand a positive half-integer mmax{1,4|z|,4|z|2}
we have
− |z| ≤ xξ≤ |z|+|z||z|
4m17
16 |z|.
Thus
ωx+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 ξξ2
2m'(2) exp(ω(m)) '(2) ex1 + y2x2
4m.
Lemma 3.
ζm'(2) exp ξξ2
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
ζmexp ξξ2
2m<5
4
|ξ|3
m2exp(ξ)
<5
417
163|z|3
m2exp x+1
16<2|z|3
m2exp (x).
Lemma 4.
exp ξξ2
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
m23
128 <1,(3.8)
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from which |eη1|<2|η|. Thus, using (3.5), (3.4), and (3.8), together with Remark 2,
exp ξξ2
2mexp (ω)
(3.5)
exp (ω)eη1
(3.4)
eη1exp 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 + y2x2
4m.
Proof. Recall that |er1r| ≤ r2(1< r < 1).With r=y2x2
4mwe have |r| ≤
|z|2
4m1
16, and therefore
exp y2x2
4m1y2x2
4m|z|4
16m2,
from which exp x+y2x2
4mex1 + y2x2
4m|z|4
16m2ex.
Theorem 2. Given z=x+iy Cwe have
δ(z, n) = ez1 + z
nn&(2)
|<(z2)|
2ne<(z).
Speciﬁcally:
δ(z, n)|<(z2)|
2ne<(z)12 |z|3
n21 + |z|
48 e<(z)
for n2 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
ζmexy2x2
4mexζmexp ξξ2
2m
+exp ξξ2
2mexp (ω)
+exexp y2x2
4m1y2x2
4m
2|z|3
m2exp(x) + |z|3
m2exp(x) + |z|4
16m2exp(x)
=M
m2
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where M= 3 |z|31 + |z|
48 exdepends only on z. Now
|ζmex| − |y2x2|
4mexζmexy2x2
4mexM
m2
and therefore
δ(z, n)(3.3)
≥ |ζmex| ≥ |y2x2|
4mexM
m2.
Remark 3. The upper bound
δ(z, n) = ez1 + z
nn|z|2
2ne|z|,
valid for zCand 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={zC:x2=y2}={zC:<(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,
ez1 + 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)mζm
elζmin
rId
dr(rm)
=melζmin exp m1
mx, ζm1.
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Now
elζ(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,
ζm1>1 + x
mm1
m
>exp m1
mxx2
2m
>exp xx
mx2
2mif m > x > 0,
while
ζm1>exp m1
mxx2
2(m+x)
>exp xx
mx2
mif m > 2x > 0.
Hence
min exp m1
mx, ζm1>exp xx
mx2
m
and therefore
δ(z, n)>|x|3
8m2exp xx
mx2
m(m > 2|x|)
|x|3
8m2ex1x(1 + x)
m.
Supremum of {δ(z, n) : zK}
Note that
δ(z, n)≤ |ez|+1 + ξ
mm
ex+eξ=ex1 + e|z|2
4mex1 + e|z|,
showing that the following is a good deﬁnition:
Deﬁnition 1. Given a bounded subset Kof Cdeﬁne
∆(K, n) = sup{δ(z, n) : zK}
for each nN.
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 n2, and uniformly over K,
∆(K, n)&(2)
1
2nsup <(z2)e<(z):zK.
If K⊂ X then, asymptotically, to within a term of order n3, and uniformly over K,
∆(K, n)&(3)
1
2n2sup |<(z)|3e<(x):zK.
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) =
ea1 + a
nn
ekak1 + kak
nn
< ekak"1exp kak2
2n!#
<kak2
2nekak.
Proof. Following [1, §8], for every nNwe 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!1n!
nk(nk)!
for every integer ksuch that 0 kn, we have
snbn=
n
X
k=0
1
k!1n!
nk
1
(nk)!
| {z }
=λk(n)0
ak=
n
X
k=0
λk(n)ak(4.3)
and
σnβn=
n
X
k=0
λk(n)kakk(4.4)
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with obviously all λk(n)0, where
λ0(n)λ1(n)0 (n1) and λ2(n)1
2n(n2) .
Using the triangle inequality we estimate
keabnk=k(easn)+(snbn)k≤keasnk+ksnbnk.
Now, considering (4.1)–(4.2),
keasnk=
X
k=n+1
ak
k!
X
k=n+1
kakk
k!=ekakσn,
and, referring to (4.3) and (4.4),
ksnbnk=
n
X
k=0
λkak
n
X
k=0
λkkakk=σnβn.
Consequently
keabnk ≤ ekakσn+ (σnβn) = ekakβn.
Therefore, from (4.1), we obtain the estimate
ea1 + a
nn
ekak1 + kak
nn
which, see Proposition 1,
< ekakekakexp kak2
2n!
and this
<kak2
2nekak
since 1 er< r (rR).
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) =
ea1 + a
nn
ea1 + a
nnσ
∆(σ(a), n)
&(2)
1
2nsup <(z2)e<(z):zσ(a)
for any a∈ A and positive integer n.
Thus, if ais not quasinilpotent,
ea1 + a
nn
tends to 0no faster than O(n1)unless
σ(a)is contained in Xin which case the rate of convergence is no faster than O(n2).
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Proof. Without loss of generality we may assume that Ais commutative. Then
σ(a) = {φ(a) : φΦ(A)}
where Φ(A) is the set of characters (nontrivial multiplicative functionals) on A.
Now, for any φΦ(A) and any n(recall that the norm dominates the spectral radius),
eφ(a)1 + φ(a)
nn=φea1 + a
nn
ea1 + a
nnσ
ea1 + 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 n1.
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 (rR): equivalently, if k1 + irhk ≤
1 + o(r) (R3r0). 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)
eh1 + h
nn
khk2
2nekhk.
When khk ∈ σ(h), then, more precisely,
khk2
2nekhk.(2)
eh1 + 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):zσ(a)≥ khk2e−khkif − khk ∈ σ(h)
and
sup <(z2)e<(z):zσ(a)≥ khk2ekhkif khk ∈ σ(h).
Remark 5. Our upper estimate (Theorem 5) for the size of ea1 + 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 ea1 + 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
 F. F. Bonsall & J. Duncan, Complete Normed Algebras, Springer (1973).
 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