JJ II

(1)

volume 5, issue 3, article 55, 2004.

Received 23 March, 2004;

accepted 01 April, 2004.

Communicated by:L.-E. Persson

Abstract Contents

JJ II

J I

Home Page Go Back

Close Quit

Journal of Inequalities in Pure and Applied Mathematics

A TWO-SIDED ESTIMATE OF e^x−(1 +x/n)ⁿ

CONSTANTIN P. NICULESCU AND ANDREI VERNESCU

Department of Mathematics University of Craiova Street A. I. Cuza 13 Craiova 200217, Romania.

EMail:[email protected] University Valahia of Târgovi¸ste Bd. Unirii 18, Târgovi¸ste 130082 Romania.

EMail:[email protected]

c

2000Victoria University ISSN (electronic): 1443-5756 062-04

(2)

A Two-Sided Estimate Of e^x−(1 +x/n)ⁿ

Constantin P. Niculescu and Andrei Vernescu

Title Page Contents

JJ II

J I

Go Back Close

Quit Page2of9

J. Ineq. Pure and Appl. Math. 5(3) Art. 55, 2004

http://jipam.vu.edu.au

Abstract

In this paper we refine an old inequality of G. N. Watson related to the formula e^x= limn→∞ 1 +^x_nn

.

2000 Mathematics Subject Classification:Primary 26D15; Secondary 26A06, 26A48 Key words: Inequalities, Two-sided estimates, Exponential function, Means.

The exponential function can be defined by the formula e^x = lim

n→∞

1 + x

n n

,

the convergence being uniform on compact subsets of R.The "speed" of convergence is discussed in many places, including the classical book of D. S.

Mitrinovi´c [2], where the following formulae are presented:

0≤e^x− 1 + x

n n

≤ x²e^x

n for|x|< nandn ∈N^? 0≤e^−x−

1− x n

n

≤ x²(1 +x)e^−x

2n for0≤x≤n, n∈N, n≥2 0≤e^−x−

1− x n

n

≤ x²

2n for0≤x≤n, n∈N^?. HereN^? stands for the set of positive naturals.

(3)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page3of9

See [3], [4], [8], [9], [10] for history, applications and related results. As noticed by G.N. Watson in [9], the first inequality yields a quick proof of the equivalence of two basic definitions of the Gamma function. In fact, forx > 0 it yields

n→∞lim Z n

0

s^x−1 1− s

n n

ds = Z ∞

0

s^x−1e^−sds,

while a small computation shows that the integral on the left is equal to n!n^x

x(x+ 1)· · ·(x+n).

The aim of the present note is to prove stronger estimates.

Theorem 1.

i) Ifx >0,t >0andt > ^1−x₂ ,then

x²e^x

2t+x+ max{x, x²} < e^x− 1 + x

t t

< x²e^x 2t+x.

ii) Ifx >0,t >0andt > ^x−1₂ ,then

x²e^−x

2t−x+x² < e^−x− 1−x

t t

< x²e^−x

2t−2x+ min{x, x²}. Forx= 1andt=n ∈N^? the inequalitiesi)yield,

e

2n+ 2 < e−

1 + 1 n

n

< e 2n+ 1,

(4)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page4of9

which constitutes Problem 170 in G. Pólya and G. Szegö [6].

Forx= 1andt=n ∈N^? the inequalitiesii)read as 1

2n e < 1 e −

1− 1

n n

< 1 (2n−1)e

and this fact improves the result of Problem B3 given at the 63rd Annual William Lowell Putnam Mathematical Competition. See [5]. Needless to say, the solu- tions presented in [1] and [11] both missed the question of whether the original pair of inequalities are optimal or not.

The result of Theorem1above can be easily extended for positive elements in a C^?-algebra (particularly in M_n(R)). This is important since the solution u∈C¹([0,∞),Rⁿ)of the differential system

(1)





 du

dt +Au= 0 fort∈[0,∞) u(0) =u₀

forA∈M_n(R),has an exponential representation, u(t) = e^−tAu₀

= lim

n→∞

I− t

nA n

u0. Sincee^−tA = e^tA−1

,we can rewriteu(t)as

(2) u(t) = lim

n→∞

"

I+ t

nA −1#n

u0.

(5)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page5of9

This led K. Yosida [7] to his semigroup approach of evolution equations:

Theorem 2. Let E be a Banach space and let A : D(A) ⊂ E → E be a densely defined linear operator such that for everyλ >0, the operatorI+λA is a bijection betweenD(A)andEwith

(I+λA)⁻¹ ≤1.

Then for every u0 ∈ D(A) the formula (2) provides the unique solution u∈C¹([0,∞), E)∩C([0,∞),D(A))of the Cauchy problem (1).

It is unclear up to what extent an analogue of Theorem 1 is valid in the context of unbounded generatorsA.

Proof of Theorem1. We shall detail here only the case i). The caseii)can be treated in a similar way.

We shall need the Harmonic, Logarithmic and Arithmetic Mean Inequality, 2uv

v+u < v−u

lnv−lnu < u+v

2 , for everyu, v >0, u < v, from which we get the following two-sided estimate

(3) 2x

2t+x <ln(t+x)−lnt < (2t+x)x

2t(t+x), for everyt, x >0.

The left-hand side inequality ini)is equivalent to (4) u(t) := 2t+x+ max{x, x²}

2t+x+ max{x, x²} −x²

1 + x t

t

<e^x fort >max

0,^1−x₂ .

(6)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page6of9

If the parameterxbelongs to(0,1],then u(t) = 2t+ 2x

2t+ 2x−x²

1 + x t

t

, so that

u⁰(t) =

ln(t+x)−lnt− x t+x

2t+ 2x

2t+ 2x−x² − 2x² (2t+ 2x−x²)²

1 + x

t t

>

2x

2t+x − x t+x

2t+ 2x

2t+ 2x−x² − 2x² (2t+ 2x−x²)²

1 + x

t t

= 2x³(1−x) (2t+ 2x−x²)²(2t+x)

1 + x

t t

≥0

by the left-hand side inequality in (3). Therefore the functionu(t)is increasing.

Aslimt→∞u(t) = e^x,this proves (4) forx∈(0,1].

Forx≥1,the inequality (4) reads u(t) = 2t+x+x²

2t+x

1 + x t

t

< e^x for everyt >0.

In this case, u⁰(t) =

2t+x+x²

2t+x − 2x² (2t+x)²

1 + x

t t

and the left part of (3) yields u⁰(t)>

2x

2t+x − x t+x

2t+x+x²

2t+x − 2x² (2t+x)²

1 + x

t t

= x³(x−1) (2t+x)²(t+x)

1 + x

t t

≥0

(7)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page7of9

sincet >0andx≥1. Thenu(t)is increasing and thus u(t)< lim

t→∞u(t) =e^x

for every t > 0and every x ≥ 1. Hence (4), and this shows that the left-hand side inequality ini)holds for everyx >0.

The right-hand side inequality in Theorem1i)is equivalent to e^x < 2t+x

2t+x−x²

1 + x t

t

=v(t) for everyx >0and everyt >max

0,^1−x₂ .Again, we shall use a monotonic- ity argument. According to the right-hand side inequality in (3) we have

v⁰(t) =

2t+x

2t+x−x² − 2x² (2t+x−x²)²

1 + x

t t

<

(2t+x)x 2t(t+x) − x

t+x

2t+x

2t+x−x² − 2x² (2t+x−x²)²

1 + x

t t

= x⁴(−2t+ (1−x))

2t(t+x)(2t+x−x²)² <0

from which we infer thatv(t)is decreasing. Consequently, v(t)> lim

t→∞v(t) =e^x for every x > 0and every t > max

0,^1−x₂ . Thus also the right-hand side inequality ini)holds and the proof is complete.

(8)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page8of9

References

[1] L.F. KLOSINSKI, The Sixty-Third William Lowell Putnam Mathematical Competition, Amer. Math. Monthly, 110(8) (2003), 718–726.

[2] D.S. MITRINOVI ´C, Analytic Inequalities, Springer-Verlag, Berlin Hei- delberg New York, 1970.

[3] E.H. NEVILLE, Note 1209: Two inequalities used in the theory of the gamma functions, Math. Gazette, 20 (1936), 279–280.

[4] E.H. NEVILLE, Note 1225: Addition to the Note 1209, Math. Gazette, 21 (1937), 55–56.

[5] C.P. NICULESCU ANDA. VERNESCU, On the order of convergence of the sequence 1− _n¹n

, (Romanian) Gazeta Matematic˘a, 109(4) (2004).

[6] G. PÓLYA AND G. SZEGÖ, Problems and Theorems in Analysis, Springer-Verlag, Berlin Heidelberg New York, 1978.

[7] K. YOSIDA, Functional Analysis, Springer-Verlag, Berlin Heidelberg New York, 7th Edition, 1995.

[8] G.N. WATSON, An inequality associated with gamma function, Messen- ger Math., 45, (1916) 28–30.

[9] G.N. WATSON, Note 1254: Comments on Note 1225, Math. Gazette, 21 (1937), 292–295.

[10] E.T. WHITTAKER AND G.N. WATSON, A Course of Modern Analysis, Cambridge Univ. Press, 1952.

(9)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page9of9

[11] *** 63rd Annual William Lowell Putnam Mathematical Competition, Math. Magazine, 76(1) (2003), 76–80.