Putnam 999 A. Find polynomials f(x),g(x), andh(x), iftheyexist, suchthatforallx, if x< f(x) g(x) + h(x) 3x + if x 0 x + if x > 0 Solution. Let h(x) ax + b. Note that there are corners in the graph of this function at x and x 0. we expect f(x) c (x +) and g(x) dx for some positive constants c and d. For x <, we have f(x) g(x) + h(x) c (x +)+dx + ax + b (d c + a) x + b c b c and d c + a 0. For x>0, x + f(x) g(x) + h(x) c (x +) dx + ax + b (c d + a) x + b + c b + c and c d + a. Now b c and b + c b,c 3 d c + a 0 and c d + a a,d c a 5 f(x) 3 (x +),g(x) 5 x, h(x) x + is correct for x< and x>0. Continuity gives 3x +for x 0. A. Let p(x) be a polynomial that is nonnegative for all real x. Prove that for some k, there are polynomials f (x),...,f k (x) suchthat p(x) kx (f j (x)). j Solution. Clearly p(x) has real coefficients, and we may assume p(x) is monic. Let p(x) r (x) c (x) where the roots of r (x) are all real and the roots of c (x) are not real. We have r(x) j c (x) k (x r j ) (x r j ) s (x) and (x ak ) + b k j
Note that when c (x) is multipied out it is a sum of squares of polynomials, say c (x) P K m g m(x), and as required. KX p(x) s (x) c (x) (s (x) g m (x)), m A3. Consider the power series expansion x x X a n x n. Prove that, for each integer n 0, there is an integer m such that n0 a n + a n+ a m. Solution. We do a partial fractions decomposition: x x (x +) (x +) +(x +) Hence, A B + x ++x ³ ³ A ++x + B x A B x x µ + x x + + µ P x n n0 ( ) n n0 µ P + + n0 a n n+ + ( ) n x n ( +) n n+ + ++x ( ) n n+ x n + + x + ( ) n n+ +
and 8 8 a n + a n+ 8 + 8 + 8 ( ) n n+ + n+ + n+ + ( )n+ n+ + n+ +( ) n + n+ + n+4 +( ) n+ + n+4 + + + + n+4 + n+ + a n+ 8 4 n+4 + n+3 + 4+ 8 n+4 + n+3 + A4. Sum the series m n Solution. Usingthefactthat P m m n m n m wherewehavealsoused m n 3 m (n3 m + m3 n ). P n P P n m,wehave m n 3 m (n3 m + m3 n ) X (3 m /m)(3 m /m +3 n /n) m n m n µ (3 m /m)(3 m /m +3 n /n) + (3 n /n)(3 n /n +3 m /m) (3 m /m)(3 n /n) X m, 3 m a (a + b) + b (a + b) b ab (a + b) + 3 a ba (a + b) ab.
Now m m 3 X m the original sum is then d dx m 3 m xm d dx m x µ x 3 x x µ 3 9 4 3. X m 3 ( 3+) 3 4 ³ x m 3 x A5. Prove that there is a constant C such that, if p(x) is a polynomial of degree 999, then Z p(0) C p(x) dx. Solution. We may assume that p(0) 6 0and that p(x) is monic. Then p(x) 999 k (x r k ) where the r k 60are the possibly complex roots, ordered by modulus. We have Q p(0) 999 R p(x) dx k r k R Q 999 k x r k dx R Q 999 k x r k dx We need to show that as a function of the r k, R Q 999 k x r k dx is bounded below. For r k >, we have x > x r k > r k > Z 999 k r k x r k dx > 999 N Z k N x dx r k where the product is over those r k with r k. For each N, f N (r,r,,r N ): Z k N x dx r k is a continuous positive function of (r,r,,r N ) { z } N, a compact set. f N has a positive minimum, say M N. Finally, Z 999 k x r k dx > min N 999 4 ª M 999 N N > 0,
We can take C min N 999 ª. M 999 N N A6. Thesequence(a n ) n is defined by a,a,a 3 4, and, for n 4, a n 6a n a n 3 8a n a n a n a n 3. Show that, for all n, a n is an integer multiple of n. Solution. a n 6a n a n 3 8a n a n a n a n 3 a n a n 6a n a n 3 8a n a n a n 3 6 a n a n 8 a n a n 3. with b n : Trying b n r n, we get an a n,wehave b n 6b n 8b n, 0 r n 6r n +8r n 0 r 6r +8 r n (r ) (r 4) r n Hence, and b n c n + c 4 n c,c 4 and b a a c + c 4 4c +6c b 3 a 3 a c 3 + c 4 3 8c +64c. b n n + n n n a n b n a n b n b n a n b n b n b a b n b n b n (n )+(n )+ + ( i ) i n n(n )/ ( i ). i We need to show that n divides a n n(n )/ Q n i (i ). Write n j k where k is odd. Certainly j n n(n )/ for n so that j divides n(n )/. It suffices to show that 5
k divides i for some i<n.we recall that since gcd (k, ), k divides φ(k) and φ (k) is less than n, as required. B. Right triangle ABC has right angle at C and BAC θ; thepointd is chosen on AB so that AC AD ;thepointe is chosen on BC so that CDE θ. The perpendicular to BC at E meets AB at F.Evaluatelim θ 0 EF. Solution. Notethat AB cos θ AC, so that The line CB has slope b : AB / cos θ secθ. sin θ b cos θ and hence its equation is cos θ sin θ sin θ cos θ cos θ cos θ sin θ cos θ sin θ cot θ Also ADC (π θ) / and y cot θ (x b) x cos θ +. sin θ EDB π θ ADC π θ (π θ) / (π θ) /. the line DE has equation µ π y tan θ µ θ cot (x ) sin θ (x ) cos θ The lines intersect when Then sin θ x cos θ + (x ) cos θ sin θ (x ) sin θ ( x cos θ +)( cos θ) (x ) ( + cos θ) x cos θ + x ( + cos θ) ( + cos θ) x cos θ + x ( + cos θ) +cosθ x +cosθ +cosθ y (cot θ) +cosθ +cosθ + sin θ. 6
Finally, EF y sin θ cot θ +cosθ sin θ +cosθ + sin θ +cos θ cos θ + + cos θ sin θ (cos θ)(+cosθ)++cosθ sin θ ( + cos θ) +cosθ 3 as θ 0. B. Let P (x) be a polynomial of degree n such that P (x) Q(x)P 00 (x), whereq(x) is a quadratic polynomial and P 00 (x) is the second derivative of P (x). Show that if P (x) has at least two distinct roots then it must have n distinct roots. Solution. Suppose that P (x) does not have n distinct roots. Then P (x) is of the form P (x) (x a) m R(x) for some m where R(a) 6 0and a C. Then P 00 (x) m (x a) m R(x)+(x a) m R 0 (x) 0 m (m ) (x a) m R(x)+m (x a) m R 0 (x)+(x a) m R 00 (x) (x a) m m (m ) R(x)+m (x a) R 0 (x)+(x a) R 00 (x) (x a) m S(x) Since S(a) m (m ) R(a) 6 0,S(x) has no factors of (x a). Then implies Q(x) c (x a) and Then or (x a) m R(x) P (x) Q(x)P 00 (x) Q(x)(x a) m S(x) R(x) cs(x) R(a) cs(a) cm (m ) R(a) c P (x) Q(x)P 00 (x) P 00 (x) m (m ) m (m ). m (m ) (x a) P 00 (x) P (x) (x a) However, by expanding P (x) in powers of (x a) we can see that this is the case only if P (x) b (x a) m for some constant b. 7
B3. LetA {(x, y) :0 x, y < }. For(x, y) A, let S(x, y) X x m y n, m n where the sum ranges over all pairs (m, n) of positive integers satisfying the indicated inequalities. Evaluate lim ( xy )( x y)s(x, y). (x,y) (,),(x,y) A Solution. Notethat X x m y n x m y n x y x y Moreover, m n m n m n m n X m n+ n x x x m y n S (x, y) x m y n x m y n + x n+ x yn + n X x m y n m X n x m y n n m+ m y m+ y xm x y n + y y y x m m x 3 y ( x)( x y) + y 3 x ( y)( y x) x y S (x, y) x y x 3 y ( x)( x y) y 3 x ( y)( y x) xy (( x y)( y x) x ( y x)( y) y ( x y)( x)) ( x)( y)( x y)( y x) xy ( x3 y 3 x + x 3 y y + x y 3 ) ( x)( y)( x y)( y x) xy ( x)( y)(xy + y + x + x y ) ( x)( y)( x y)( y x) xy xy + y + x + x y ( x y)( y x), 8
and lim ( (x,y) (,),(x,y) A xy )( x y)s(x, y) 3. B4. Letf be a real function with a continuous third derivative such that f(x),f 0 (x),f 00 (x),f 000 (x) are positive for all x. Suppose that f 000 (x) f(x) for all x. Showthatf 0 (x) < f(x) for all x. Solution. Let c lim x f(x) 0. By replacing f(x) by f(x) c we may assume c 0. Let c 0 lim x f 0 (x) 0. Thenf 0 (x) >dand for x<0 f (0) f(x)+ Z 0 x f 0 (x)dx > f (x)+c 0 x. f (x) <f(0) c 0 x and it follows that c 0 0. Similarly, c 00 lim x f 00 (x) 0.Sincef 000 (x) f(x), Also, as we then have f 00 (x)f 000 (x) f 00 (x)f(x) <f 00 (x)f(x)+f 0 (x) or 0 < d dx f 0 (x)f(x) f 00 (x). lim f 0 (x)f(x) f 00 (x) c 0 c x (c00 ) 0, f 0 (x)f(x) > f 00 (x). d 3 f(x) 3f(x)f 0 (x) >f 0 (x)f(x)+ f 0 (x)f 000 (x) dx 4 f 00 (x) + f 0 (x)f 000 (x) > d dx f 0 (x)f 00 (x) or d 3 dx 4 f(x) f 0 (x)f 00 (x) > 0. Since lim 3 x 4 f(x) f 0 (x)f 00 (x) 3 4 c c0 c 00 0,wethenhave 3 4 f(x) f 0 (x)f 00 (x) > 0. Hence, or 3 4 f(x) f 0 (x) > f 0 (x) f 00 (x) d 4 dx f(x) 3 > 6 d dx f(x) 3 > 3 f 0 (x) 3 f 0 (x) 3. 9
and f 0 (x) < 3 /3 f(x) < f(x). B5. For an integer n 3, letθ π/n. Evaluate the determinant of the n n matrix I + A, wherei is the n n identity matrix and A (a jk ) has entries a jk cos(jθ + kθ) for all j, k. Solution. Letz e iθ.then cos (mθ) e imθ + e imθ z m + z m z m + z m. Let v (z, z,z 3,,z n ) and v (z, z, z 3,, z n ). The matrix A is half the sum of the outer products v v and v v. I + A I + (v v + v v) For any w C n, (I + A) w w + ((v w) v +(v w) v) where v w is the usual bilinear dot product. For u C n orthogonal to both v and v, we have (I + A) u u. This shows that I +A is the identity on span (v, v). we need only evaluate det(i + A) on span (v, v). Notethat (I + A)(v) v + ((v v) v +(v v) v) + (v v) v + (v v) v (I + A)(v) v + ((v v) v +(v v) v) (v v) v + + (v v) v the matrix of I + A on span (v, v) is + (v v) (v v) (v v) + (v v) and the determinant is (since z n and z n ) + (v v) n n + (v v) + (z + + z n ) n n + (z + + z n ) ³ + z zn n z ³ n + z zn z n n n. 4 0
B6. LetS be a finite set of integers, each greater than. Suppose that for each integer n there is some s S such that gcd(s, n) or gcd(s, n) s. Show that there exist s, t S such that gcd(s, t) is prime. Solution. Let p <p <... < p k be the set of all of the primes that occur in at least one of the prime factorizations of the elements of S. S S be the subset consisting of all those members of S involving p in their prime factorization. If there are s, t S with gcd(s, t) p, then we are done. If not, let p i p and let p i >p be the smallest prime that occurs in the factorization of members of S S, and let S S S be the subset consisting of all those members of S involving p i in their prime factorization. If there are s, t S with gcd(s, t) p i, then we are done. We continue in this way obtaining p i <p i <...<p ij where the process stops at some j k. Note that we either have s, t S j with gcd(s, t) p ij (in which case we are done) or S j ms m. Suppose that S j ms m.letn p i p i p ij. Then by assumption, there is some s S, such that gcd (s, n) s or gcd (s, n). Now gcd (s, n) isimpossible, since S j ms m. gcd (s, n) s, andsos is a divisor of n p i p i p ij.letj 0 be the smallest integer so that there is a divisor s S of p i p i p ij 0. Now p ij 0 must occur (exactly once) as a factor of s or else j 0 can be reduced. Select s j 0 S j 0. We claim gcd (s, s j 0)p ij 0. Indeed, by definition s j 0 has factor p ij 0, but no factors of p i,p i, p ij 0.