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A Maximum Principle for High-Order DerivativesAuthor(s): David PanSource: The American Mathematical Monthly, Vol. 120, No. 9 (November), pp. 846-848Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/10.4169/amer.math.monthly.120.09.846 .
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1/4 with u = 9, x = 1/4 or u = 4, x = 1/9, and our old acquaintance, the area 1/7,with u = 4, x = 1/2, or u = 2, x = 1/4, or u = 1/2, x = 4; see Figures 7 and 8.
We conclude this note with an invitation for the reader to find a purely geometricargument along the lines of [8, p. 9] for the claims indicated in the last two figures.
REFERENCES
1. R. J. Cook, G. V. Wood, Feynman’s triangle, Mathematical Gazette 88 (2004) 299–302.2. H. S. M. Coxeter, Introduction to Geometry, second edition. Wiley, New York, 1969.3. B. Curgus, The theorems of Ceva and Menelaus: an animation, available at http://faculty.wwu.edu/
curgus/Papers/Monthly2012.html.4. J. S. Kline, D. Velleman, Yet another proof of Routh’s theorem, Crux Mathematicorum 21 (1995) 37–40.5. H. Nakamura, K. Oguiso, Elementary moduli space of triangles and iterative processes, The University of
Tokyo Journal of Mathematical Sciences 10 (2004) 209–224.6. I. Niven, A new proof of Routh’s theorem, Math. Mag. 49 (1976) 25–27, available at http://dx.doi.
org/10.2307/2689876.7. E. J. Routh, A Treatise on Analytical Statics with Numerous Examples, Vol. 1, second edition. Cambridge
University Press, London, 1909, available at http://www.archive.org/details/texts.8. H. Steinhaus, Mathematical Snapshots, third English edition. Translated from the Polish. With a preface
by Morris Kline. Dover Publications, Mineola, NY, 1999.
Department of Mathematics, Western Washington University, Bellingham, WA 98225, USA,[email protected]
Department of Mathematics, Western Washington University, Bellingham, WA 98225, USA,[email protected]
A Maximum Principle forHigh-Order Derivatives
David Pan
Abstract. We prove a maximum principle for high-order derivatives under initial conditions.
1. INTRODUCTION We begin with a well-known definition.
Definition 1.1. The maximum principle holds for f (x) on [a, b] if f (x) ≤ max{ f (a),f (b)} on [a, b].
It is well known that f ′(x) ≥ 0 implies the maximum principle because of nonde-creasingness, and that f ′′(x) ≥ 0 implies the maximum principle because of noncon-cavity. It is also well known that f (n)(x) ≥ 0 where n ≥ 3 does not necessarily implythe maximum principle. For example, consider f (x) = ±x2 and its third derivative.
We present conditions under which f (n)(x) ≥ 0 where n ≥ 3 does imply the maxi-mum principle. Let I = [a, b] be a closed interval of the real line, and let Cn(I ) be the
http://dx.doi.org/10.4169/amer.math.monthly.120.09.846MSC: Primary 26A06
846 c© THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 120
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set of real-valued functions on I that are n-times continuously differentiable. Recallthat
Pn(x) :=n∑
i=0
f (i)(a)
i !(x − a)i
is the nth-degree Taylor polynomial of f (x) at a, where f (x) ∈ Cn(I ) and a ∈ I .
Theorem 1.1. Let f (x) ∈ Cn([a, b]) for some n ≥ 2. If f (n)(x) ≥ 0 on [a, b], thenf (x) ≤ vn(x) on [a, b], where
vn(x) := Pn−2(x)+f (b)− Pn−2(b)
(b − a)n−1(x − a)n−1.
The following theorem is a corollary of Theorem 1.1. It shows that the maximumprinciple holds under initial conditions.
Theorem 1.2. Let f (x) ∈ Cn([a, b]) for some n ≥ 3. If
(i) f (n)(x) ≥ 0 on [a, b], and
(ii) f (b) ≥ Pi (b) for i = 1, . . . , n − 2,
then f (x) ≤ max{ f (a), f (b)} on [a, b] (i.e., the maximum principle holds for f (x)on [a, b]).
Example 1.1. Below is a graph of f (x) = (x − 2)(x − 1)(x + 1)(x + 2) = x4−
5x2+ 4.
−3 −2 −1 1 2 3
10
20
30
40
It is false that f ′(x), f ′′(x) ≥ 0 on [0, 3], but it is true that f ′′′(x) ≥ 0 on [0, 3].Also, f (3) = 40 ≥ P1(3) = 4. Therefore, by Theorem 1.2, the maximum principleholds for f (x) on [0, 3].
Proof of Theorem 1.1. Suppose that f (x) ∈ Cn([a, b]) for some n ≥ 2 and f (n)(x) ≥0 on [a, b]. By Taylor’s theorem,
g(x) :=f (x)− Pn−2(x)
(x − a)n−1=
1
(n − 2)!
∫ 1
0(1− t)n−2 f (n−1)(a + t (x − a)) dt.
November 2013] NOTES 847
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It follows that
g′(x) =1
(n − 2)!
∫ 1
0t (1− t)n−2 f (n)(a + t (x − a)) dt.
Since f (n)(x) ≥ 0 on [a, b], g′(x) ≥ 0 on [a, b], so g(x) ≤ g(b) on [a, b], whichimplies
f (x) ≤ Pn−2(x)+f (b)− Pn−2(b)
(b − a)n−1(x − a)n−1
= vn(x)
on [a, b].
Proof of Theorem 1.2. We prove Theorem 1.2 by mathematical induction.First, we prove the initial case n = 3. Suppose that for some f (x) ∈ C3([a, b]), we
have f ′′′(x) ≥ 0 on [a, b] and f (b) ≥ P1(b). By Theorem 1.1, we have
f (x) ≤ v3(x) = P1(x)+f (b)− P1(b)
(b − a)2(x − a)2
on [a, b]. Because P ′′1 (x) = 0 and f (b)− P1(b) ≥ 0, we have v′′3 (x) ≥ 0, which, byconvexity, implies that v3(x) ≤ max{v(a), v(b)} on [a, b]. It is obvious that v3(a) =f (a) and v3(b) = f (b), so f (x) ≤ v3(x) ≤ max{ f (a), f (b)} on [a, b]. Therefore, thetheorem holds in the initial case n = 3.
Now, we assume that the theorem holds in the case n = k where k ≥ 3, and wewant to show that the theorem holds in the case n = k + 1. To do this, suppose thatfor some f (x) ∈ Ck+1([a, b]), we have f (k+1)(x) ≥ 0 on [a, b] and f (b) ≥ Pi (b) fori = 1, . . . , k − 1. By Theorem 1.1, we have
f (x) ≤ vk+1(x) = Pk−1(x)+f (b)− Pk−1(b)
(b − a)k(x − a)k
on [a, b]. Because P (k)k−1(x) = 0 and f (b)− Pk−1(b) ≥ 0, we have v(k)k+1(x) ≥ 0. It is
obvious that v(i)k+1(a) = f (i)(a) for i = 0, . . . , k − 1 and vk+1(b) = f (b), so
vk+1(b) = f (b) ≥ Pj (b) =j∑
i=0
f (i)(a)
i !(b − a)i =
j∑i=0
v(i)k+1(a)
i !(b − a)i
for j = 1, . . . , k − 2. By the induction assumption,
vk+1(x) ≤ max{vk+1(a), vk+1(b)} = max{ f (a), f (b)}
on [a, b]. Therefore, f (x) ≤ vk+1(x) ≤ max{ f (a), f (b)} on [a, b]. This concludesthe mathematical induction, and the proof is complete.
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