volume 5, issue 3, article 55, 2004.
Received 23 March, 2004;
accepted 01 April, 2004.
Communicated by:L.-E. Persson
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Journal of Inequalities in Pure and Applied Mathematics
A TWO-SIDED ESTIMATE OF ex−(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:cniculescu@central.ucv.ro University Valahia of Târgovi¸ste Bd. Unirii 18, Târgovi¸ste 130082 Romania.
EMail:avernescu@pcnet.ro
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2000Victoria University ISSN (electronic): 1443-5756 062-04
A Two-Sided Estimate Of ex−(1 +x/n)n
Constantin P. Niculescu and Andrei Vernescu
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Abstract
In this paper we refine an old inequality of G. N. Watson related to the formula ex= limn→∞ 1 +xnn
.
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 ex = lim
n→∞
1 + x
n n
,
the convergence being uniform on compact subsets of R.The "speed" of con- vergence is discussed in many places, including the classical book of D. S.
Mitrinovi´c [2], where the following formulae are presented:
0≤ex− 1 + x
n n
≤ x2ex
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.
A Two-Sided Estimate Of ex−(1 +x/n)n
Constantin P. Niculescu and Andrei Vernescu
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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
sx−1 1− s
n n
ds = Z ∞
0
sx−1e−sds,
while a small computation shows that the integral on the left is equal to n!nx
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
x2ex
2t+x+ max{x, x2} < ex− 1 + x
t t
< x2ex 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,
A Two-Sided Estimate Of ex−(1 +x/n)n
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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 Mn(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∈Mn(R),has an exponential representation, u(t) = e−tAu0
= lim
n→∞
I− t
nA n
u0. Sincee−tA = etA−1
,we can rewriteu(t)as
(2) u(t) = lim
n→∞
"
I+ t
nA −1#n
u0.
A Two-Sided Estimate Of ex−(1 +x/n)n
Constantin P. Niculescu and Andrei Vernescu
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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
<ex fort >max
0,1−x2 .
A Two-Sided Estimate Of ex−(1 +x/n)n
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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) = ex,this proves (4) forx∈(0,1].
Forx≥1,the inequality (4) reads u(t) = 2t+x+x2
2t+x
1 + x t
t
< ex for everyt >0.
In this case, u0(t) =
ln(t+x)−lnt− x t+x
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
A Two-Sided Estimate Of ex−(1 +x/n)n
Constantin P. Niculescu and Andrei Vernescu
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sincet >0andx≥1. Thenu(t)is increasing and thus u(t)< lim
t→∞u(t) =ex
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 ex < 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) =
ln(t+x)−lnt− x t+x
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) =ex for every x > 0and every t > max
0,1−x2 . Thus also the right-hand side inequality ini)holds and the proof is complete.
A Two-Sided Estimate Of ex−(1 +x/n)n
Constantin P. Niculescu and Andrei Vernescu
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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.
A Two-Sided Estimate Of ex−(1 +x/n)n
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[11] *** 63rd Annual William Lowell Putnam Mathematical Competition, Math. Magazine, 76(1) (2003), 76–80.