f (t) dt Is Constant, Must f (t) = c/t? This work is motivated by the calculus problem of finding the derivative of 2x x 1 t dt, x 0

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1 If F() = f (t) dt Is Constant, Must f (t) = c/t? Tian-Xiao He, Zachariah Sinkala, and Xiaoya Zha Tian-Xiao He (the@iwu.edu) received his Ph.D. from Teas A&M University in 99 and is currently a professor of mathematics at Illinois Wesleyan University (Bloomington, IL 670). His research interests include approimation theory, spline function theory, wavelet analysis, and computational combinatorics. He is particularly interested in multivariate splines, spline wavelets, numerical approimation, and generating functions. Zachariah Sinkala (zsinkala@mtsu.edu) earned his Ph.D. from the University of South Florida in 989. He is a professor of mathematics at Middle Tennessee State University (Murfreesboro, TN 373). His research interests include partial differential equations, both theory and applications in the life sciences. He is particularly interested in protein dynamics (proton transport in proteins). Xiaoya Zha (zha@mtsu.edu) is an associate professor of mathematics at Middle Tennessee State University (Murfreesboro, TN 373). He received his Ph.D. from Ohio State University in 993. His research interests include graph theory and combinatorics, and he is particularly interested in topological and geometric aspects of graphs. Introduction This work is motivated by the calculus problem of finding the derivative of F() = t dt, 0 (a problem designed for applying the Fundamental Theorem of Calculus and the Chain Rule). It is easy to see that F () = 0, which implies a nice geometric fact: for any given 0, the area between the curves y = /t, y = 0, and the -ais from to is constant. Clearly, for any constant c, the function y = c/t also has this property. It is natural to ask whether the converse is true or not; that is, if f is a continuous function on R\{0} and if F, f () = f (t) dt is a constant function of ( 0), must f (t) = c/t for some constant c R? The eample in the net section shows that the answer is no. Similarly, there eists a function g c/ for which F 3,g () = 3 g(t) dt is a constant function of. However, one may ask whether both F,h () and F 3,h () being constant functions leads to h(t) = c/t? Equivalently, does there eist a function k(t) c/t that can be constructed by combining f and g so that both F,k () and F 3,k () are constant? We study this problem in the net section and VOL. 36, NO. 3, MAY 005 THE COLLEGE MATHEMATICS JOURNAL 99

2 give an answer in the negative; that is, if both F,k () and F 3,k () are constant, then k(t) must be a constant multiple of /t. In addition, we will show that if positive reals p and q satisfy ln p/ ln q / Q, then both F p, f () and F q, f () are constants if and only if f (t) = f ()/t. The study of the problem relies on dense subsets of R. Inthe net section we give a collection of small subsets that are dense in R. (Only after finishing the paper did we find out that a special case of Theorem in the net section was obtained by Moser and Macon []. They derived a similar result for the integer setting. Moreover, Heuer also gave an alternative proof of Moser and Macon s result in []. However, we have decided to retain the section as we generalize the result to pairs of real numbers and give a necessary and sufficient condition. In addition, our proof is different, self-contained, and elementary.) Our main results will be shown in the net section, and their proofs are given in the final section. Main results For simplicity, we restrict our attention to functions defined on R + ; clearly our results and functions can be etended to hold on R {0}. Let f (t) be a continuous function on R +,andlet F λ, f () = λ f (t) dt, λ R + \{}. If f (t) = c/t, c R, then for > 0, F λ, f () = c ln λ, a constant. However, the following eample shows that the converse is not true; that is, F λ, f () being constant does not imply that f (t) = c/t for some constant c. Eample. Defineafunction f (t) recursively (see Figure ): (i) for t [, ), let [ t, 3 ) t, f (t) = t +, t [ ) 3, ; (ii) for t [ n, n ), n =, 3,...,let f (t) = (/ n ) f (t/ n ); (iii) for t [/ n, / n ), n =,,...,let f (t) = n f ( n t). Then f (t) is continuous on R + and F, f () = f (t) dt = 4. In fact, this eample can be generalized to a function f λ (t) with any parameter λ in a natural way. For λ>, define f λ (t) on R + as follows: for t [λ n,λ n ), n Z,let f λ (t) = λ n (t λn ), λ n (t λn ) t t [ λ n, ) λn (λ + ) [ ) λn (λ + ), λ n. 00 c THE MATHEMATICAL ASSOCIATION OF AMERICA

3 Figure. Figure of function f (t) when t [ 4, /] [/, 4] For 0 <λ<, define f λ (t) on R + similar to the case λ>, with the only change being replacing λ by /λ. The function f λ (t) is continuous on R +,and F λ, fλ () = λ f λ (t) dt = 4 (λ ), a constant. Clearly, f λ (t) c/t for any c R. On the other hand, it is not hard to show (by the same approach as in the proof of Theorem, e.g., choosing n = andn = 3) that if f (t) is a continuous function on R + and F n, f () = n f (t) dt is a constant function for all positive integers n, then f (t) and a function c/t agree on all rational numbers. Hence by the continuity of f, f (t) = f ()/t. Certainly, the condition that, for all n Z +, the function F n, f () is constant is very strong. The following theorem shows that in fact we can weaken the sufficient condition considerably. Theorem. Let p and q be positive reals with ln p/ln q irrational, and let function f (t) be continuous on R +.Then f(t) = f ()/t if and only if both F p, f () and F q, f () are constant on R +. The complete proof will be given in net section, but we outline it here. Define S p,q ={±p k q l : k, l Z}. We first show that for every t S p,q, f (t) = f ()/t; that is, for all integers k and l, f (p k q l ) = f ()/(p k q l ). Thus, if S p,q is dense in R +,then f () = f ()/ for all R +. It is easy to see that S p,q is not dense in R + if ln p/ ln q is rational. However, the following theorem shows that when ln p/ ln q is irrational, S p,q is dense in R +. Theorem. Let p and q be positive reals different from.thens p,q is dense in R + if and only if log q p is irrational. VOL. 36, NO. 3, MAY 005 THE COLLEGE MATHEMATICS JOURNAL 0

4 Proofs To establish Theorem, we need a technical lemma. Lemma. Let p and q be positive reals different from with log q p / Q.Then (i) There eists a decreasing sequence {a n } of numbers in S p,q for which lim a n = + ; n (ii) There eists an increasing sequence {b n } of numbers in S p,q for which lim n b n =. Proof. Since S p,q is closed under taking reciprocals, we may assume that < p < q. Lets = ln p and t = ln q. Then0< s < t. Since ln p is not a rational multiple of ln q, there eists an integer n Z + such that Let n s < t <(n + )s and t n s + (n + )s. s = min{t n s,(n + )s t } and We have () 0 < s < s < t < s < t (since t = ma{t n s,(n + )s t }. t n s + (n + )s and n s < t <(n + )s ); () t is not a rational multiple of s (otherwise (n + )s t = k(t n s ), k Q, which implies that t is a rational multiple of s, i.e., ln p ln q (= log q p) Q, a contradiction); (3) s and t are linear combinations of ln p and ln q with integer coefficients. Construct s, t, s 3, t 3,...inductively. Assume s i and t i are constructed, satisfying (4) 0 < s i < s i < t i < s i < t i ; (5) t i is not a rational multiple of s i ; (6) s i and t i are linear combinations of ln p and ln q with integer coefficients. Let s i+ = min{t i n i s i,(n i + )s i t i } and t i+ = ma{t i n i s i,(n i + )s i t i }, where n i is the unique integer satisfying n i s i < t i <(s i + )s i. It is easy to show inductively that (4) (6) are true for s i+ and t i+. Therefore, two sequences {s n, n } and {t n, n } are constructed, satisfying (4) (6). By (4), s > s > > s n > > 0 0 c THE MATHEMATICAL ASSOCIATION OF AMERICA

5 and lim n s n = 0 +.Forn =,,...,let a n = e s n.then a n S p,q, a > a > > a n > >, and lim n a n = +. Thus part (i) of the lemma is true. Let b n = /a n, n. Then b < b < < b n < < and lim n b n =. Therefore part (ii) of the lemma is also true. Proof of Theorem. Let c R +.Ifc S p,q, there is nothing to prove, so suppose c / S p,q. We want to find a sequence { n } in S p,q with n c.lets = S p,q (0, c) and α = sup S.Thenα c, and there is a sequence { n } in S converging to α,soit suffices to show that α = c. We shall show that α<c is impossible. If α<c we may by the lemma choose a i in S p,q such that < a i < + (c α)/α. Thus α/a i <α, so there eists n 0 such that α/a i < n0 <α.then n0 a i S p,q but ( α< n0 a i <αa i <α + c α ) α = α + c < c, showing that n0 a i S and n0 a i >α, which is impossible because α = sup S. Thus α = c. Proof of Theorem. The necessity of the given condition is obvious, so we need only prove the sufficiency. The function f (t) is continuous, and hence F p, f () is differentiable over R\{0}. SinceF p, f () is a constant function of, wehavef p, f () = pf(p) f () = 0, andinturn f (p) = f ()/p. Inductively, for l Z + and l >, we have f (p l ) = f (pl ) p = = f () p l. Furthermore, f () = f ( p ) = f ( ) p p p, and therefore f (/p) = pf(). Inductively we have f (/p n ) = p n f (). Similarly, for k Z +, f (q k ) = f () ( ) and f = q k f (). q k q k It follows that f (p k q l ) = f (), k, l Z. p k ql Let S p,q.then =±p k q l for all k, l Z.If = p k q l,then f () = f (p k q l ) = f () = f (). p k q l VOL. 36, NO. 3, MAY 005 THE COLLEGE MATHEMATICS JOURNAL 03

6 If = p k q l,then f () = f (p k q l ( )) = f ( ) p k q l = f ( ). If is a positive real number, there eists a sequence of numbers { k : k S p,q } such that lim k k =.Since f is continuous on R\{0}, f () f () = lim f ( k ) = lim = f (). k k k If is negative, a similar argument shows that f () = f ( )/. Since F p, f () is a constant function of, p f ( ) t dt = p f () t dt. Evaluating the above integrals, we obtain f ( ) ln p = f () ln p. Thus f ( ) = f (), completing the proof. The authors would like to thank the referees and the editor for their sug- Acknowledgments. gestions and help. References. G. A. Heuer, Rational numbers generated by two integers, Amer. Math. Monthly 78 (97) L. Moser and N. Macon, On the distribution of first digits of powers, Scripta Math. 6 (950) W. Rudin, Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, 976. Numeration Nine figures are sufficient to epress any number in common practice: nevertheless, the following table may be thought necessary. Nonillions Octillions Septillions Setillions Quintillions 85734, 6486, , 43796, 4347, Quadrillions Trillions Billions Millions Units 4806, 354, 6734, 36849, The American Tutor s Assistant Revised; or, A Compendious System of Practical Arithmetic Printed by Joseph Crukshank, Philadelphia, 809, page. 04 c THE MATHEMATICAL ASSOCIATION OF AMERICA