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6462Author(s): J. LeVeque, Lloyd N. Trefethen and James C. SmithSource: The American Mathematical Monthly, Vol. 92, No. 10 (Dec., 1985), pp. 740-741Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2323240 .
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740 PROBLEMS AND SOLUTIONS [December
+ +Xd, but m / - 9(B). Then x =(xl,..., d) A. Let us prove .9(A) = _9(Zd). If xn E .9(Zd), then n = y2 + * +y y Let y =(y.... , Yd).
If y E A or x + y E A, then
Vn = IIy - Oil = II(x + y) - xli is in 9(A). Otherwise, y E B and x + y E B, whence
Fm = JI(x + y)- yll E- -9( B), a contradiction.
Also solved by C. Hurd, A. Kovacec (Austria), G. S. Lessells (Ireland), 0. P. Lossers (The Netherlands), M. D. Meyerson, N. Miku (The Netherlands), D. Moews, Y. Peres and N. Kremerman (Israel), D. A. Rawsthorne, D. Richman, A. Rosenfeld, D. K. Skilton, and N. C. Wormald (Australia), H. G. Williams, and the proposer.
ADVANCED PROBLEMS
Solutions of these Advanced Problems should be mailed in duplicate to Professor D. H. Mugler, Department of Mathematics, University of Santa Clara, Santa Clara, CA 95053, by April 30, 1986. The solver's full post-office address should be on each sheet.
6505. Proposed by Harry Gonshor, Rutgers University.
Prove that for any function f: R -- R there is a function g which differs from f on a set of measure 0 and has the Darboux (i.e., intermediate value) property but is not continuous.
6506. Proposed by Ronald Evans, University of California, San Diego.
Let p > 3 be prime and write f(n) for the Legendre symbol (n/p). Prove that p-l
E f(m3 + 2)(1 -f(4m + 5)) = 1. m=O
SOLUTIONS OF ADVANCED PROBLEMS
An Inequality for Rational Functions
6462 [1984, 371]. Proposed by J. LeVeque and Lloyd N. Trefethen, Courant Institute of Mathematical Sciences, New York University.
Let S be the unit circle (or any other circle) in the complex plane. Let r be a rational function of type (n, n) with no poles on S.
(a) Show that llr'lll < 4,rnjjrjI. where 11 11 and 11 Iloo are the L1 and Loo norms on S. Note that llr'll is the arclength of the image of S under r.
(b)* Show that llr'll < 2,rnjjrjj0. Solution of part (a) by James C. Smith, University of South Alabama, Mobile, Alabama. Take
S to be the unit circle and let k = max{ n, m }. We will show that, if r is a rational function of type (n, m) with no poles on S, then
llr'll < 2(zk + n + m)lrlII..
When m = n the bound 2n(v + 2)IIrII. is better than 4vrnIIr I I. We have r = P/Q where P and Q are polynomials of degree n and m, respectively. Let
f(O) = Ir(e'0)l and g(O) = arg(r(e'0)). Then f2(O) = r(e'0)r(e'0), where r(e'0) is a rational
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1985] PROBLEMS AND SOLUTIONS 741
function in e-'@ of type (n, m). It is easy to check that e'(n-m)o df2 can be expressed as A/B, where A and B are polynomials in ei? and the degree of A does not exceed 2(n + m). Thus
-d 2 2f d =o dO f = 2tdOf= 0
for at most 2(n + m) values of 0 in [0, 2 m). It follows that, for some a, [ a, a + 2'r) is the union of at most 2( n + m) disjoint half-open intervals on which f( 0) is monotone and nonnegative, and hence that
IIf'111 = j+2 If'l < 2(n + m) Ilfl.
Note that r can be written as the product of linear fractional transformations say Tl, T2,... , Tk. Since each Ti maps S one to one and onto a circle, we have
| | d arg( T,) 2Xr. dO
Now, g(O) = Earg(Ti) and so ljg'111 < 27Tk. In conclusion, since r(e i) = f(O)eig@), we have
r' = ie'gfg' + eigf'
Hence
lIr'll S Ilfg'lll + lif'III < 2,gkilf I10 + 2(n + m)lIfII1O = 2n(vr + 2)IIrII,,.
The proposers also solved part (a). Part (b)* remains open.
A Least Upper Bound
6464 [1984, 440]. Proposed by Roger Cooke, University of Vermont.
What is the least upper bound of real numbers b such that there exists a continuous real-valued function f(x) satisfying
xf(x) - X| (f(t) +tf(t)2)dt - bIn(x) oo asx > oo?
Solution by The University of South Alabama Problem Group, Mobile, Alabama. Set h(x) =
xf(x) + (2). Then we must find the least upper bound L of real numbers b satisfying
h(x) - 1tdt -(b - - ln(x) ---oo as x -+oo.
If h(x) 0 and b < 4, then the limit is indeed infinite, so L > 4. On the other hand we will
show that h (x) - j ht7 dt takes on non-positive values for x arbitrarily large.
Suppose, to the contrary, that there exists a continuous function h(x) and an xo > 1 such that for x > xo,
h(x) - j h(t)2 dt > 0.
jx h(t)2 for >Hx0x= H'(x) 1 Set H(x) = dt, then for x > xo, h(x X) xH'(x) >-. Integrating both
1 t H(x) X
sides of this inequality from xo to x, we have
H( x) _'- H(x) > In(x) - ln(xo).
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