Content uploaded by Philip G Spain

Author content

All content in this area was uploaded by Philip G Spain on May 31, 2017

Content may be subject to copyright.

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)

vl & pgs 3May 31, 2017

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