Vol. 10 No 7 - Pi Mu Epsilon

586 PI MU EPSILON JOURNAL
PROBLEMS AND SOLUTIONS 587
= I. Therefore, the average of the eigenvectors is no less than 0 nor more
than I; the upper bound being achieved by the identity matrix. The
hypotheses that A is real and symmetric are not needed.
II. Solution by Bob Prielipp, University of Wisconsin-Oshkosh, Oshkosh,
Wisconsin.
By hypothesis A is an n x n idempotent matrix. Therefore, the trace of
A equals the rank r of A. (See Exercise 4(ii) on p. 239 with solution of p.
426 of Daniel T. Finkbeiner II, Introduction to Matrices and Linear
Transformations, 3rd ed., W. H. Freeman, I978.) Since the sum of the
eigenvalues of A is equal to the trace of A and r ~ n, then the average of the
eigenvalues equals rln ~ I.
Also solved by Paul S. Bruckman, Highwood, IL, Russell Euler and Jawad Sadek, Northwest
Missouri State University, Maryville, Joe Howard and John Jeffries, New Mexico Highlands
University, Las Vegas, Murray S. Klamkin, University of Alberta, Canada, Henry S. Lieberman,
Waban, MA, David E. Manes, SUNY College at Oneonta, Kandasamy Muthuvel, University of
Wisconsin-Oshkosh, William H. Peirce, Delray Beach, FL, Henry J. Ricardo, Medgar Evers College,
Brooklyn, NY, H.-J. Seiffert, Berlin, Germany, and the Proposer.
893. [Fall I 996] Proposed by Peter A. Lindstrom, Irving, Texas.
Show that the sequence {xn} converges and find its limit, where x 1 = 2
and, for n ~ I,
2x 11
sinx + sinx + cosx X = II II
11
11+l
2 smx .
11
I. Solution by Michael K Kinyon, Indiana University South Bend, South
Bend, Indiana.
Ifj{x) = e·.r(sinx + cosx), then observe that
-x
_ e "(sinx 11
+ cosx 11
) f(x 11
)
X 11
+l - X 11
- = X - --.
-2e -x,. sinx
11
Thus the recurrence is just the Newton-Raphson method applied to the
function f. If it converges at all, it must converge to one of the zeroes of J,
which are (k + 3/4)1t, k an integer. In this case the method starts at 2 and
converges to 37t/4. Note thatf(x)f''(x) > 0 on the open interval joining the
initial point 2 to the appropriate zero of J, a sufficient condition for
11
/
1
(X11 )
convergence. Here f(x)f''(x) = -2e-2.r cos 2x and is positive on the interval
(7t/4, 37t/4), which contains 2. . . . ·-
It is perhaps somewhat intellectually dishonest to pretend to cleverly pull
f out of thin air. Observing the recursion formula in the form
xll+l = x,. -
sinx 11
+ cosx 11
2sinx 11
I guessed the problem had been constructed by application of the Newton­
Raphson method to a function of the form j{x) = g(x)(sinx + cosx). Then
differentiatej{x) and substitute into the equation
j{x) = sinx + cosx
f'(x) 2sinx
Equating coefficients yields the differential equations
g'(x) - g(x) = -2 and g'(x) + g(x) = 0.
g(x)
g(x)
The solution to the second equation is g(x) = ce·.r for some constant c, which
also satisfies the first equation. I took c = I.
II. Solution by Lisa M Croft, Messiah College, Grantham, Pennsylvania.
The recursion equation can be written in the form
so define
X X + .! + !cotx 11
,
n+l = 11
2 2
1 1
g(x) = x + -
2 2
+ -cotx.
Now g(x) and g'(x) are continuous and 0 ~ g'(x) ~ 1/2 on the interval [ap]
= [7t/4, 37t/4]. Furthermore a ~ g(x) ~ b on [a, b]. Then the contraction
mapping theorem (Elementary Numerical Analysis by Kendall Atkinson, p.
84) guarantees a unique point a. such that a.= g(a.) in the interval [a, b]. In
addition, for any initial x 0
in that interval, the iterates xn will converge to _a..
We find that fixed point by solving the equation x = g(x), which reduces .. to
cotx = -I, so x = 37t/4. Hence the sequence {xn} converges to 37t/4.