PROBLEMS AND SOLUTIONS - Cezar Lupu · 2011-03-02 · PROBLEMS AND SOLUTIONS EDITORS Curtis Cooper CMJ Problems Department of Mathematics and Computer Science University of Central

PROBLEMS AND SOLUTIONSAuthor(s): Curtis Cooper and Shing S. SoSource: The College Mathematics Journal, Vol. 40, No. 2 (March 2009), pp. 131-141Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/25653688 .

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PROBLEMS AND SOLUTIONS EDITORS

Curtis Cooper CMJ Problems Department of Mathematics and Computer Science

University of Central Missouri

Warrensburg, MO 64093

email: [email protected]

Shing S. So CMJ Solutions Department of Mathematics and Computer Science

University of Central Missouri

Warrensburg, MO 64093

email: so @ ucmo. edu

This section contains problems that challenge students and teachers of college mathematics. We urge

you to participate actively by submitting solutions and by proposing problems that are new and inter

esting. To promote variety, the editors welcome problem proposals that span the entire undergraduate curriculum.

Proposed problems should be sent to Curtis Cooper, either by email as a pdf, T^X, or Word

attachment (preferred) or by mail to the address provided above. Whenever possible, a proposed prob lem should be accompanied by a solution, appropriate references, and any other material that would

be helpful to the editors. Proposers should submit problems only if the proposed problem is not under

consideration by another journal. Solutions to the problems in this issue should be sent to Shing So, either by email as a pdf,

T|=X, or Word attachment (preferred) or by mail to the address provided above, no later than June 15, 2009.

896, Proposed by Darij Grinberg (student), Ludwig Maximilian University, Munchen,

Germany and Cezar Lupu (student), University of Bucharest, Bucharest, Romania.

Let a, b, c be positive reals such that a + b + c = 3. Show that

897. Proposed by R. S. Tiberio, Wellesley High School, Wellesley, ME.

Let G be the centroid, O the circumcenter, and N the Nagel point of a scalene triangle ABC. Show that OG _L ON if triangle ABC contains an angle of 60?.

898. Proposed by Mihdly Bencze, Sacele, Brasov, Romania.

Suppose / is a function defined on an open interval / such that f"(x) > 0 for all x e /, and [a, b] c /. Prove that

PROBLEMS

f f(a + (b-a)y)dy >

VOL. 40, NO. 2, MARCH 2009 THE COLLEGE MATHEMATICS JOURNAL 131



899. Proposed by Ovidiu Furdui, University of Toledo, Toledo, OH.

Let k be a natural number. Prove that

2 fl Xn ^ 1 hm n /- dx ? k >

-

Jo l+xk+x2k+ ---+xnk ?1(1+km)2 m=0

900. Proposed by Michel Bataille, Rouen, France.

Show that for p > 0,

where//n-ELii

SOLUTIONS

Pythagorean-triple solutions

871. Proposed by Greg Oman, Ohio State University, Columbus, Ohio.

Determine all positive integers n for which the equation x(x + n) = y2 has a solution in positive integers x and y.

Solution by Daniele Degiorgi, Massagno, Switzerland; Gerald Heuer, Concordia Col

lege, Moorhead, MN; and Joel Hams, University of North Dakota, Grand Forks, ND

(independently).

For fixed n and y, we have

-ft + Jn2 + Ay2 X

=-2-'

which is not an integer unless n2 + (2y)2 is a square, say z2. Therefore, (n, 2y, z) must be a Pythagorean triple, and hence

n = p2 ?

q2, y = pq, and z = p2 + q2

where p and q are positive integers with p > q. It follows that

x = -p2 + q2 + p2 + q2 2

Thus, for every positive integer ft expressible as a difference of two squares, the equa tion x(x + n) = y2 has positive integer solutions. Consider the following four cases:

Case 1. If n ? 1, 2, or 4, + 4j2 is not a square for if n2 = z2 ?

4j2 =

(z ?

2y)(z + 2y), then n2 is the product of two integers of the same parity. Clearly, l2 and 22 are not. As for n = 4, 42 = 2 8, but z - 2y = 2 and z + 2y = 8 imply z = 5 and y = |.

Gzs^ 2. If ft is odd and n > 3, then

n = 2fc + 1 = (Jk + l)2 - k2 for A: G N,

which is the difference of two squares, and hence the solution exists.

132 ? THE MATHEMATICAL ASSOCIATION OF AMERICA



Case 3. If n = 2k such that k is odd, then Case 2 guarantees the equation x*(x* +

k) = y*2 has a solution in positive integers x* and y*. Letting x = 2x* and y = 2y*, it follows from

that the solution exists.

Case 4. If n = 4k for any positive integer k such that k > 2, then ? = (2i)(2j) for positive integers i and y such that / > j > 1. Letting jc ? (/

? y)2 and j = (/ +

7)0' _

j), it follows that

x(x +n) = (i - j)2[(i -

j)2 + = (i - j)2(i + j)2 = y2.

Therefore, the given equation has positive integer solutions for any positive integer n ̂ 1,2, or 4.

Also solved by Armstrong Problem Solvers, Armstrong Atlantic State U.; Dionne Bailey, Elsie

Campbell, and Charles Diminnie (jointly), Angelo State U.; Herb Bailey, Rose-Hulman Inst. Tech.; Michel Bataille, Rouen, France; Brian Beasley, Presbyterian C; Tom Beatty, Florida Gulf Coast

U.; Kenneth Boback, Pennsylvania State U., Wilkes-Barre Campus; Brian Bradie, Christopher Newport U.; James Carpenter, Iona C; John Christopher, California State U., Sacramento; Margaret Cibes,

Trinity C, Hartford, CT; Elliott Cohen, Fontenay-sous-Bois, France; Robert Crise, Crafton Hills C; Chip

Curtis, Missouri Southern State U.; Matthew Davis (student), Thomas Peter, and Xiaoshen Wang

(jointly), U. of Arkansas, Little Rock; Michael DiPasquale (student), Wheaton C, IL; James Duemmel,

Bellingham, WA; Arthur Foss, Plantation, FL; Russell Hendel, Towson U.; Hofstra University

Problem Solvers, Hofstra U.; Peter Hohler, Aarburg, Switzerland; Houghton College Problem

Solving Group, Houghton C; Rebecca Jacobs (student), Harvard-Westlake School; Tom Jager, Calvin

C; Lenny Jones, Shippensburg U.; Panagiotis T. Krasopoulos, Athens, Greece; Elias Lampakis,

Kiparissia, Greece; Bryce Lampe and Andrew Bernoff (jointly), Harvey Mudd C; Manuel Lopez, IES

Maria Sarmiento, Viveiro, Spain; John Noonan, Mount Vernon Nazarene U.; Northwestern University

Math Problem Solving Group, Northwestern U.; Michael Parmenter, Memorial U. of Newfoundland, St, John's, NL, Canada; Paul Peck, Glenville State C; John Perry and Ashley Sanders (jointly), U. of Southern Mississippi; Kevin Roper, Cedarville U.; Christopher Schafhauser (student), U. of

Wisconsin-Platteville; William Seaman, Albright C; Harry Sedinger, St. Bonaventure U.; Seton Hall

University Problem Solving Group, Seton Hall U.; The Skidmore College Problem Group, Skidmore C; David Sluss, C.C. of Allegheny County; Philip Straffin, Longmont, CO; John Sumner and AlDA Kadic-Galeb (jointly), U. of Tampa; james swenson, U. of Wisconsin-Patteville; H.T. tang,

Hayward, CA; Bob Tomper, U. of North Dakota; Daniel Vacaru, Colegiul National "Zinca Golescu",

Pite?ti, Romania; Michael Vowe, Therwil, Switzerland; Gary Walls, West Texas A & M U.; George

Wene, U. of Texas at San Antonio; Konstantine Zelator, U. of Toledo; hongbiao Zeng, Fort Hays State

U.; John Zerger, Catawba C; and the proposer.

Two incorrect solutions were received.

872. Proposed by Jose Luis Diaz-Barrero, Universidad Politecnica de Cataluna, Barcelona, Spain.

Prove that, for every positive n,

x(x +n) = 2x\2x* + 2k) = 4[x*(jc* + it)] = (2j*)2 = y2

A lower bound for 1

(nl)2

E

n

(n-k)\(n + k)\ 2(-\)k ~

(n + l)(2n + 1) > 6




Solution by Eugene Herman, Grinnell College, Grinnell, IA.

We will derive the sharper lower bound 7/(ft2 + 5n + 1). Note that

7 6 n2 + 5n + 1

~ (ft + l)(2n + 1)

for all n > 1,

which is readily confirmed by cross-multiplying and expanding. We can simplify the

given sum by using the identity

?<-<)-<-? (V) (Gould, Combinatorial Identities, formula 1.86):

= J_y:2(_ir-,YM

= ^(2?-1)

= ̂

We must therefore prove that

2/?

for all ? > 1 ft2 + 5ft + 1

which is equivalent to

ft(ln(ft2 + 5ft + l)-ln7) > 21nft! for all ft > 1. (1)

Inequality (1) is clearly true for n = 1 and may be confirmed numerically for n ?

2, 3,... ,8. Thus, from now on, we assume n > 9. We use Taylor's remainder formula

1 9 ln(l + x) > x-x when 0 < x < 1

2

and Stirling's approximation

y- / 1\ 1 In ft! < In V27T + \ n + -

)mn ?n + ?

\ 2/ 6ft

(Feller, An Introduction to Probability and its Applications, 2nd ed., page 52). Thus, when ft > 9,

ft(ln(ft2 + 5ft + 1) - In7) = n (inn2 +

In ^1

+ ^ + -

In7^ /5 1 1/5 1 \2\

> 2ftlnft + ft - +

? -- - +

? -win7

\ ft ft2 2 \/i ft2/ /

1 25 f ̂ = 2ft In ft + 5 H-(

? + ? ) - ft In 7

ft 2ft

23 > 2ft lnft + 4.93 - ? - 1.95ft

2ft




and

1 2\nn\ < \n2n +2nlnn + \nn -2n + ?.

3n

Hence, to prove (1), it suffices to prove

1 23 ln27r +2nlnn + lnn - 2n + ? < 2nlnn +4.93 -- l.95n

3n 2n

which simplifies to

12

n + In n < 0.05n + 4.93 - In 2;r for all n > 9. (2)

This may be confirmed numerically for n = 9. Let

12 f(x) = 0.05x-In x +4.93 - ln2jr.

x

Then

12 1 f(x) = 0.05 + ? - - > 0 for x > 9.

X2 X

We see that

12 1 - + - < 0.05 for all n > 9 n2 n

which proves (2) and hence (1).

Note: By Stirling's formula, (n\)~2/n ̂ e2 In2, which suggests a lower bound of the form a/(n2 + bn + c). Since \e2\

? 7, this is the largest possible integer value of a. The proposer used a = 3. For sufficiently large values of n, we could use b = 2 instead of b = 5. Since |"ln27r] = 2, this is the smallest possible integer value of b.

Also solved by Jemal Mohammed-Awel, Chunlei Liu, and Charles Kicey (jointly), Valdosta State U.; Dionne Bailey, Elsie Campbell, Charles Diminnie, and Roger Zarnowski (jointly), Angelo State

U.; Michel Bataille, Rouen, France; Brian Bradie, Christopher Newport U.; Margaret Cibes, Trinity C, Hartford, CT; Chip Curtis, Missouri Southern State U.; James Duemmel, Bellingham, WA; Hofstra

University Problem Solvers, Hofstra U.; Tom Jager, Calvin C; Harris Kwong, SUNY Fredonia; Elias Lampakis, Kiparissia, Greece; Masao Mabuchi, Hachinohe National C. of Tech.; Northwestern University Math Problem Solving Group, Northwestern U.; Rob Pratt, Raleigh, NC; William

Seaman, Albright C; John Sumner and Aida Kadic-Galeb (jointly), U. of Tampa; Bob Tomper, U. of North Dakota; Daniel Vacaru, Colegiul National "Zinca Golescu", Pite?ti, Romania; Michael Vowe, Therwil, Switzerland; Albert Whitcomb, Castle Shannon, PA; and the proposer.

Editor's note. Dionne Bailey, Elsie Campbell, Charles Diminnie, and Roger Zarnowski of Angelo State University; Hofstra University Problem Solvers; Tom Jager of Calvin

College; and Harris Kwong of SUNY Fredonia independently sharpen the result in

problem 872 as follows:

2(-l)* xU"

y_ > *

f^{n-k)\{n+k)\ ~

(n + l)2




Stieltjes constants 873. Proposed by Ovidiu Furdui, University of Toledo, Toledo, Ohio.

Ink Find (a) ?, and (b) ̂(-l)*"1

\n2k

k=\

Solution by John Sumner and Aida Kadic-Galeb, University of Tampa, Tampa, FL; and Bob Tomper, University of North Dakota, Grand Forks, ND (independently).

More generally, for integers n > 0, the sum

In" A:

k=\

lrf_k\ _ ln"+1 m

can be expressed in terms of the Stieltjes constants defined by

Yn = lim n?>cx

In particular,

Yo = lim

is the well-known Euler constant. In fact, we can establish the following generalization: for any nonnegative integer

n,

E k=\

^ (-ly^ln^ First, consider the partial sum S2m of / -;- Then

k=\

->2m

2m

E k=l

(-\y-llnnk

^\nnk 2^\nn2k 2s ~k 2^ 2k k=\

2m \nnk

k=\

Elll

k (In 2 + In k)n

'glnÂ ln"+12m n + 1

2m \nnk\ \nn+l2m

E U=i

n + 1

m i n / \

k=\ K j=0 \J/

lnn+1 2m

n + 1




Rewrite the last term above as

i=0 V/ k=\ K i=0 v-// 7=0

Then

E u=i

lnÂ hV+1m hV+1m

}~ j + l + 7 + 1

lim 52m = V m?vnn ? (?1)*?1 In" A:

k=i

-?0 y/ln"--/2+ lim

m-ôo

lim m^-oo

lim

lim ra?>oo

'lnw+12m

w + 1

lnn+12m

n + 1

J=o\J

+i

7 + 1

n-\

1^0 \J

lnn+12m (ln2 + lnmf+1 ln"+12 +

In

n + 1

n + 1 50

n + 1

Yj lnn"; 2.

n + 1

n-\

j=0

Therefore,

E (-I)*"'In it lnz2

(?l)*-1 In2 In3 2

k=\

(ln2)y0 and

(ln22)Ko-2(ln2)yi.

Also solved by arkady Alt, San Jose, CA; michel bataille, Rouen, France; Brian Bradie, Christopher Newport U.; Hongwei Chen, Christopher Newport U.; Donal Connon, Kent, England; G.R.A.20 Prob lem Solving Group, Rome, Italy; Manuel Lopez, IES Maria Sarmiento, Viveiro, Spain; Paolo Perfetti,

Dipartimento di Matematica, Universita degli studi di Tor Vergata Roma, Rome, Italy; William Seaman,

Albright C; nora Thornber, Raritan Valley C.C; Michael Vowe, Therwil, Switzerland; and the proposer. A computer generated solution was submitted by Tony Tam (student), San Diego State U.

An infinite sum with f(4n>(1) and ^2n>(1) 874. Proposed by Michael S. Becker, University of South Carolina at Sumter, Sumter, South Carolina.

Let f{x) = 1/(1 + x2), and let f(n)ix) denote its rath derivative. Show that

~ //(* )(!) /<2">(1)\ =

fÂ (4n)! Qn)\ ) "




Solution by by Ovidiu Furdui, University of Toledo, Toledo, OH

We will prove more generally that for a natural number k > 1 the following equality holds:

oo^ fi2kn){X) ^ ^ I 2k - cos kf + sin k-f

n=0 (2kn)\ 22k _ 2*+i Cos kf + 1

Since -40 with/ ^T, the nth derivative of / is given by

/(->(,) = <zl> (_!_L_\ J 2i V(^-0"+1 (jc + i)"+1) _ (-1)"?! /(x + 0"+1 ~ ~2i V 0^+ 1)

(* -

o n+i

n+1

and hence

{2nk)\

1 2i (^r-(?)

2nit+l'

Since 1 + i = -J2e '*

and 1 ? i = \f2e '* , we can deduce that

[_ ?=0 N / n=0 N '

Ink

22k

22k - (1 + 02V 2 V22*-(i-02V

2* - cos ^ + sin 2

22* - 2k+l cos ^ + 1

*7T

When jfc = 1 and 2,

^ (2?)! 5 > -= -

respectively,

and the result follows.

Also solved by Herb Bailey and roger Lautzenheiser (jointly), Rose-Hulman Inst. Tech.; michel

Bataille, Rouen, France; Brian Bradie, Christopher Newport U.; Hongwei Chen, Christopher Newport

U.; Elliott Cohen, Fontenay-sous-Bois, France; Charles Diminnie, Angelo State U.; James Duemmel,

Bellingham, WA: Michael Goldenberg and Mark Kaplan (jointly), Baltimore Poly. Inst.; G.R.A.20

Problem Solving Group, Rome, Italy; Eugene Herman, Grinnell C; Hofstra University Problem

Solvers, Hofstra U.; Tom Jager, Calvin C; Harris Kwong, SUNY Fredonia; Kee-Wai Lau, Hong

Kong, China; Chunlei Liu, Valdosta State U.; masao Mabuchi, Hachinohe National C. of Tech.; KlM

McInturff, Santa Barbara, CA; Jerry Minkus, San Francisco, CA; Northwestern University Math

Problem Solving Group, Northwestern U.; Paolo Perfetti, Dipartimento di Matematica, Universita degli studi di Tor Vergata Roma, Rome, Italy; William Seaman, Albright C; Peter Simone, U. of Nebraska

Medical Center; John Sumner and Aida Kadic-Galeb (jointly), U. of Tampa; James Swenson, U.

of Wisconsin-Platteville; Bob Tomper, U. of North Dakota; Daniel Vacaru, Colegiul National "Zinca

Golescu", Pitesti, Romania; Michael Vowe, Therwil, Switzerland; Albert Whitcomb, Castle Shannon,

PA; John Zacharias, Melbourne, FL; and the proposer.




Editor's note. Jerry Minkus of San Francisco, CA has also established the following

generalization of Problem 874: Let f(x) = 1/(1 + jc2), and let f(n)(x) denote its nth

derivative. Then

^\ (4n)\ (2n)\ ) (a4 + 3a2 + 6 a2 + 2

+ + 4)(a2+4) a4+4

Two cyclic inequalities 875. Proposed by Dorin Marghidanu, Colegiul National "A. I. Cuza," Corabia, Romania.

Let a, b, c, and d be positive real numbers such that a+b + c + d = 2. Prove that

(a) y/a(b + c) + V&(c + d) + + a) + >Mtf + &) < 3, and

(b) \6yjabcd(a + b)(b + c)(c + d)(d + a) < (2 - a)(2 - 6)(2 - c)(2 - d).

Solution 1 by Harris Kwong, SUNY Fredonia, Fredonia, NY.

We shall extend both results to positive real numbers a, b, c, and d whose sum is S.

For (a), we obtain a tighter inequality by applying the Cauchy-Schwartz inequality:

y/a(b + c) + + d) + y/c(d + a) + ̂(a + fc)

< Vfl + * + c + V2(a + b + c + d) = V2S.

For (b), it follows from the Arithmetic-Geometric Mean Inequality that

2y/d(b + c)<(b + c) + d = S-a and 2yjb(c + d) < b + (c + d) = S - a.

Therefore,

4y/bd(b + c)(c + d) < (S - a)2,

4y/ca(c + d)(d + a) < (S - b)2,

4y/db(d + a)(a+b) < (S - c)2, and

4jac(a + b)(b + c) < (S - d)2.

It follows that

I6y/abcd(a + b){b + c)(c + + a) < (S - a)(S - b)(S - c)(S - d).

Solution 2 for part (b) by Chip Curtis, Missouri Southern State University, Joplin, MO and Kee-Wai Lau, Hong Kong, China (independently).

By the Arithmetic-Geometric Mean Inequality, we can establish

2 ? a = b + c + d

_ 3 ̂ (fe + c) + (c + d) + (d + b)^

> l[(b + c)(c + d)(d + b)}x/\




Similar inequalities can be established for 2 ? b, 2 ? c, and 2 ? d. It follows that

(2 -

fl)(2 -

Z>)(2 -

c)(2 -

</)

> ?^[(a + fc)(a + c)(a + d)(b + c)(fe + d)(c + d)]2/3 lo

= ^V(a + b)(b + c)(c + d){d + a)[(a + b)(b + c)(c + d)(d + a)]l/6 16

[(a + c)(b + d)]2/3 81 > ?Via + b)(b + c)(c + d){d + a)[(2Vab)(2Vb^)(2V^d)(2Vd^],/6 16

[(2V^)(2VM)]2/3 81 , = ?yjabcd(a + b)(b + c)(c + d)(d + a),

and hence the result follows.

Also solved by Michel Bataille, Rouen, France; Berry College Dead Poets Society, Berry C; Brian Bradie, Christopher Newport U.; Elsie Campbell, Dionne Bailey, Charles Diminnie, and Roger Zarnowski Oointly), Angelo State U.; Minh Can, Irvine Valley C; Margaret Cibes, Trinity C, Hartford, CT; Elliott Cohen, Fontenay-sous-Bois, France; Gordon Crandall, LaGuardia CC; Richard Crow (student), California State U., Fullerton; James Duemmel, Bellingham, WA; I.J.L. Garces, Ateneo de Manila U., The Philippines; David Getling, Christchurch, New Zealand; Michael Goldenberg and Mark Kaplan (jointly), Baltimore Poly. Inst.; Hofstra University Problem Solvers, Hofstra U.; Tom

Jager, Calvin C; Satyajit Karmakar, Gordon C; Paul Kominers, Massachusetts Inst. of Tech. and Scott Kominers, Harvard U. (students, jointly); Yu-Ju Kuo, Indiana U. of Pennsylvania; Elias Lampakis,

Kiparissia, Greece; Wen Liu (student), Kent State U.; Manuel Lopez, IES Maria Sarmiento, Viveiro, Spain; Northwestern University Math Problem Solving Group, Northwestern U.; Paolo Perfetti,

Dipartimento di Matematica, Universita degli studi di Tor Vergata Roma, Rome, Italy; Daniel Poore (student), Pomona C; William Seaman, Albright C; Seton Hall University Problem Solving Group, Seton

Hall U.; John Sumner and Aida Kadic-Galeb (jointly), U. of Tampa; George Tsapakidis, Kypseli Aitoloakarnanias, Greece; Daniel Vacaru, Colegiul National "Zinca Golescu", Pite?ti, Romania; Michael

Vowe, Therwil, Switzerland; Albert Whitcomb, Castle Shannon, PA; Hongbiao Zeng, Fort Hays State

U.; and the proposer.

Editor's note. The inequality in Problem 875(b) can be further generalized as follows: Let a, b, c, and d be positive real numbers such that a+b + c-\-d = S. Then

Editor's note. Costas Efthimiou of University of Cental Florida noted that in order to

rectify the confusion caused by the free index / in the solution of the generalized result of CMJ Problem 854 at the top of page 245 in Vol. 39, No. 3 (2008), one may proceed as follows: First, rewrite

? y/abcd(a + b)(b + c)(c + d)(d + a) < (S - a)(S - b)(S - c)(S - d).

in the form




The following infinite sum can be written in terms of n + 1 iterated integrals as

follows.

E k=\ k(k+\)--.(k + n)

r r rx"+* p

in

Jo Uo Uo

r*2 ln(l -x{)

id -xx)

Applying a well-known identity to the right-hand side above

dxn dxn+\.

E k=\ k(k + 1) ...(k + n) n\ JQ t(\-t)

For x = 1, the integral of the published solution can be obtained:

Hk If1 (1 - f)"-' ln(l - 0 E

k=\ k(k+l)...(k + - =

?( n) n\ J0 dt.

By replacing the free index / with an n, the term xk+n on the left-hand side at the

top of page 245 can be considered as xk+lxn~l. Following the steps in the published solution, the required result can be established. Note that the extra x11 is irrelevant.

I also felt confident that my career as a mathematician would not be adversely affected by university discipline, arrests, or any notoriety I might get from being part of an SDS action. I had read about the history of mathematics and was aware of the long tradition of tolerance of eccentricity and political dissidence among mathematicians.

The lieutenant had reached the right conclusion, but for the wrong reason. Flying a helicopter that had been repaired by a communist would not necessarily have been a mistake. However, flying one that had been repaired by a mathematician would definitely have been a bad idea.

?Neal Koblitz, Random Curves

(reviewed on page 142)




Documents

PROBLEMS AND SOLUTIONS - Cezar Lupu · 2011-03-02 · PROBLEMS AND SOLUTIONS EDITORS Curtis Cooper CMJ Problems Department of Mathematics and Computer Science University of Central