Solutions to Calculus problems. b k A = limsup n. a n limsup b k,

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1 Solutions to Calculus problems. We will prove the statement. Let {a n } be the sequence and consider limsupa n = A [, ]. If A = ±, we can easily find a nondecreasing (nonincreasing) subsequence of {a n } with the limit + ( ). Suppose A R. If infinitely many of the a n s are equal to A, we can take them as the (both nondecreasing and nonincreasing) subsequence. Otherwise, there are either infinitely many a n s larger than A or infinitely many a n s smaller than A. Suppose there are infinitely many of a n s, forming a subsequence {b k }, larger than A (the other case is analogous). Then lim inf k b k A = limsup n a n limsup b k, k so lim b k = A, b k > A. Now we can easily construct a nonincreasing subsequence {b kl } of {b k } by induction: Take k 1 = 1, i. e. b k1 = b 1. If k l is defined, we have b kl > A and limb k = A, so there exist an index k l+1 > k l such that b kl+1 < b kl. Since all b k s, including just chosen b kl+1, are larger than A, we can proceed..[1996-a1] If x and y are the sides of two squares with combined area 1, then x + y = 1. Suppose without loss of generality that x y. Then the shorter side of a rectangle containing both squares without overlap must be at least x, and the longer side must be at least x + y. Hence the desired value of A is the maximum of x(x + y). To find this maximum, we let x = cosθ, y = sinθ with θ [, π/4]. Then we are to maximize cos θ + sin θ cosθ = 1 (1 + cosθ + sin θ) = 1 + cos(θ π/4) 1 +, with equality for θ = π/8. Hence this value is the desired value of A. 6.[1998-B1] Notice that (x + 1/x) 6 (x 6 + 1/x 6 ) (x + 1/x) 3 + (x 3 + 1/x 3 ) = (x + 1/x) 3 (x 3 + 1/x 3 ) = 3(x + 1/x) (difference of squares). The latter is easily seen (e.g., by AM-GM) to have minimum value 6 (achieved at x = 1). 7.[1998-A3] If at least one of f(a), f (a), f (a), or f (a) vanishes at some point a, then we are done. Hence we may assume each of f(x), f (x), f (x), and f (x) is either strictly positive or strictly negative on the real line. By replacing f(x) by f(x) if necessary, we may assume f (x) > ; by replacing f(x) by f( x) if necessary, we may assume f (x) >. (Notice that these substitutions do not change the sign of f(x)f (x)f (x)f (x).) Now f (x) > implies that f (x) is increasing, and f (x) > implies that f (x) is convex, so that f (x + a) > f (x) + af (x) for all x and a. By letting a increase in the latter inequality, we see that f (x + a) must be positive for sufficiently large a; it follows that f (x) > for all x. Similarly, f (x) > and f (x) > imply that f(x) > for all x. Therefore f(x)f (x)f (x)f (x) > for all x, and we are done. 8.[1998-B3] We use the well-known result that the surface area of the sphere cap {(x, y, z) x +y +z = 1, z z } is simply π(1 z ). (This result is easily verified using calculus; we omit the derivation here.) Now the desired surface area is just π minus the surface areas of five identical halves of sphere caps; these caps, up to isometry, correspond to z being the distance from the center of the pentagon to any of its sides, i.e., z = cos π 5. Thus the desired area is π ( 5 π(1 cos π 5 )) = 5π cos π 5 3π (i.e., B = π/). 9.[4-B3] The answer is {a a > }. If a >, then the function f(x) = a/(a ) has the desired property; both perimeter and area of R in this case are a /(a ). Now suppose that a, and let f(x) be 1

2 a nonnegative continuous function on [, a]. Let P = (x, y ) be a point on the graph of f(x) with maximal y-coordinate; then the area of R is at most ay since it lies below the line y = y. On the other hand, the points (, ), (a, ), and P divide the boundary of R into three sections. The length of the section between (, ) and P is at least the distance between (, ) and P, which is at least y ; the length of the section between P and (a, ) is similarly at least y ; and the length of the section between (, ) and (a, ) is a. Since a, we have y + a > ay and hence the perimeter of R is strictly greater than the area of R. 1.[6-B5] The answer is 1/16. We have x f(x)dx with equality when f(x) = x/. xf(x) dx = (x 3 /4 x(f(x) x/) )dx x 3 /4 dx = 1/16, 11.[5-B3] First solution: The functions are precisely f(x) = cx d for c, d > arbitrary except that we must take c = 1 in case d = 1. To see that these work, note that f (a/x) = dc(a/x) d 1 and x/f(x) = 1/(cx d 1 ), so the given equation holds if and only if dc a d 1 = 1. If d 1, we may solve for a no matter what c is; if d = 1, we must have c = 1. (Thanks to Brad Rodgers for pointing out the d = 1 restriction.) To check that these are all solutions, put b = log(a) and y = log(a/x); rewrite the given equation as f(e b y )f (e y ) = e b y. Put then the given equation rewrites as g(y) = log f(e y ); or g(b y) + log g (y) + g(y) y = b y, log g (y) = b g(y) g(b y). By the symmetry of the right side, we have g (b y) = g (y). Hence the function g(y) + g(b y) has zero derivative and so is constant, as then is g (y). From this we deduce that f(x) = cx d for some c, d, both necessarily positive since f (x) > for all x. Second solution: (suggested by several people) Substitute a/x for x in the given equation: Differentiate: Now substitute to eliminate evaluations at a/x: f (x) = a xf(a/x). f a (x) = x f(a/x) + a f (a/x) x 3 f(a/x). f (x) = f (x) x + f (x) f(x). Clear denominators: xf(x)f (x) + f(x)f (x) = xf (x). Divide through by f(x) and rearrange: = f (x) f(x) + xf (x) xf (x) f(x) f(x).

3 The right side is the derivative of xf (x)/f(x), so that quantity is constant. That is, for some d, Integrating yields f(x) = cx d, as desired. f (x) f(x) = d x. 1.[-B3] The desired inequalities can be rewritten as 1 1 ( (1 n < exp + n log 1 1 )) < 1 1 n n. By taking logarithms, we can rewrite the desired inequalities as ( log 1 1 ) ( < 1 n log 1 1 ) ( < log 1 1 ). n n n Rewriting these in terms of the Taylor expansion of log(1 x), we see that the desired result is also equivalent to 1 i i n i < 1 (i + 1)n i < 1 in i, which is evident because the inequalities hold term by term. Note: David Savitt points out that the upper bound can be improved from 1/(ne) to /(3ne) with a slightly more complicated argument. (In fact, for any c > 1/, one has an upper bound of c/(ne), but only for n above a certain bound depending on c.) 13.[3-A3] First solution: Write f(x) = sin x + cosx + tanx + cotx + secx + csc x 1 sinx + cosx = sin x + cosx + + sin xcosx sin xcos x. We can write sin x + cosx = cos(π/4 x); this suggests making the substitution y = π/4 x. In this new coordinate, and writing c = cosy, we have sin xcosx = 1 sin x = 1 cosy, ( f(y) = (1 + c) 1 + ) c 1 = c + 1 c 1. We must analyze this function of c in the range [, ]. Its value at c = is 3 <.4, and at c = is + 3 > 6.4. Its derivative is 1 /(c 1), which vanishes when (c 1) =, i.e., where c = 1 ±. Only the value c = 1 is in bounds, at which the value of f is 1 > As for the pole at c = 1, we observe that f decreases as c approaches from below (so takes negative values for all c < 1) and increases as c approaches from above (so takes positive values for all c > 1); from the data collected so far, we see that f has no sign crossings, so the minimum of f is achieved at a critical point of f. We conclude that the minimum of f is 1. Alternate derivation (due to Zuming Feng): We can also minimize c + /(c 1) without calculus (or worrying about boundary conditions). For c > 1, we have 1 + (c 1) + c

4 by AM-GM on the last two terms, with equality for c 1 = (which is out of range). For c < 1, we similarly have c + 1 c 1 +, here with equality for 1 c =. Second solution: Write f(a, b) = a + b + 1 ab + a + b ab. Then the problem is to minimize f(a, b) subject to the constraint a + b 1 =. Since the constraint region has no boundary, it is enough to check the value at each critical point and each potential discontinuity (i.e., where ab = ) and select the smallest value (after checking that f has no sign crossings). We locate the critical points using the Lagrange multiplier condition: the gradient of f should be parallel to that of the constraint, which is to say, to the vector (a, b). Since and similarly for b, the proportionality yields f a = 1 1 a b 1 a a b 3 a 3 b + a 3 b 3 + a b =. The irreducible factors of the left side are 1+a, 1+b, a b, and ab a b. So we must check what happens when any of those factors, or a or b, vanishes. If 1 + a =, then b =, and the singularity of f becomes removable when restricted to the circle. Namely, we have f = a + b + 1 a + b + 1 ab and a + b 1 = implies (1 + b)/a = a/(1 b). Thus we have f = ; the same occurs when 1 + b =. If a b =, then a = b = ± / and either f = + 3 > 6.4, or f = 3 <.4. If a =, then either b = 1 as discussed above, or b = 1. In the latter case, f blows up as one approaches this point, so there cannot be a global minimum there. Finally, if ab a b =, then a b = (a + b) = ab + 1 and so ab = 1 ±. The plus sign is impossible since ab 1, so ab = 1 and f(a, b) = ab + 1 ab + 1 = 1 > This yields the smallest value of f in the list (and indeed no sign crossings are possible), so 1 is the desired minimum of f. Note: Instead of using the geometry of the graph of f to rule out sign crossings, one can verify explicitly that f cannot take the value. In the first solution, note that c+/(c 1) = implies c c+ =, which has no real roots. In the second solution, we would have Squaring both sides and simplifying yields a b + ab + a + b = 1. a 3 b 3 + 5a b + 4ab =, whose only real root is ab =. But the cases with ab = do not yield f =, as verified above. 4

5 14.[-A4] We use integration by parts: sin xsin x sin x dx = x sin x (xdx) = sin x B ( x cosx cosx + x sin x ) x cosx dx. Now sin x x cosx tends to as B, and the integral of sin x x cosx converges absolutely by comparison with 1/x. Thus it suffices to note that cosx x cosx dx = cosx 4x cosx (xdx) = cosx 4x sin B x xcosx sin x 4x 3 sin x dx, and that the final integral converges absolutely by comparison to 1/x 3. An alternate approach is to first rewrite sinxsin x as 1 (cos(x x) cos(x + x). Then cos(x + x)dx = x + 1 sin(x + x) converges absolutely, and cos(x x) can be treated similarly. B sin(x + x) (x + 1) dx 15.[5-A5] First solution: We make the substitution x = tanθ, rewriting the desired integral as Write π/4 log(tan(θ) + 1) dθ. log(tan(θ) + 1) = log(sin(θ) + cos(θ)) log(cos(θ)) and then note that sin(θ) + cos(θ) = cos(π/4 θ). We may thus rewrite the integrand as 1 log() + log(cos(π/4 θ)) log(cos(θ)). But over the interval [, π/4], the integrals of log(cos(θ)) and log(cos(π/4 θ)) are equal, so their contributions cancel out. The desired integral is then just the integral of 1 log() over the interval [, π/4], which is π log()/8. Second solution: (by Roger Nelsen) Let I denote the desired integral. We make the substitution x = (1 u)/(1 + u) to obtain I = = = log() (1 + u) log(/(1 + u)) (1 + u ) log() log(1 + u) 1 + u du du 1 + u I, du (1 + u) yielding I = 1 log() du 1 + u = π log(). 8 5

6 Third solution: (attributed to Steven Sivek) Define the function f(t) = log(xt + 1) x + 1 so that f() = and the desired integral is f(1). Then by differentiation under the integral, By partial fractions, we obtain whence and hence f (t) = dx x (xt + 1)(x + 1) dx. f (t) = t arctan(x) log(tx + 1) + log(x + 1) (t + 1) = πt + log() 4 log(t + 1) 4(t, + 1) f(t) = log()arctan(t) f(1) = π log() 4 + π log(t + 1) 8 log(t + 1) t dt. + 1 x=1 x= t log(t + 1) t dt + 1 But the integral on the right is again the desired integral f(1), so we may move it to the left to obtain and hence f(1) = π log()/8 as desired. Fourth solution: (by David Rusin) We have log(x + 1) x + 1 f(1) = π log() 4 dx = ( ) ( 1) n 1 x n n(x dx. + 1) We next justify moving the sum through the integral sign. Note that ( 1) n 1 x n dx n(x + 1) is an alternating series whose terms strictly decrease to zero, so it converges. Moreover, its partial sums alternately bound the previous integral above and below, so the sum of the series coincides with the integral. Put x n dx J n = x + 1 ; then J = arctan(1) = π 4 and J 1 = 1 log(). Moreover, J n + J n+ = x n dx = 1 n + 1. Write A m = m ( 1) i 1 i 1, B m = m ( 1) i 1 ; i 6

7 then Now the N-th partial sum of our series equals N J n 1 n 1 J n n = N J n = ( 1) n (J A n ), J n+1 = ( 1) n (J 1 B n ). ( 1) n 1 n 1 (J 1 B n 1 ) ( 1)n n (J A n ) = A N (J 1 B N 1 ) + B N (J A N ) + A N B N. As N, A N J and B N J 1, so the sum tends to J J 1 = π log()/8. Remarks: The first two solutions are related by the fact that if x = tan(θ), then 1 x/(1 + x) = tan(π/4 θ). The strategy of the third solution (introducing a parameter then differentiating it) was a favorite of physics Nobelist (and Putnam Fellow) Richard Feynman. Noam Elkies notes that this integral is number.491#8 in Gradshteyn and Ryzhik, Table of integrals, series, and products. The Mathematica computer algebra system (version 5.) successfully computes this integral, but we do not know how. 7

The 66th William Lowell Putnam Mathematical Competition Saturday, December 3, 2005

The 66th William Lowell Putnam Mathematical Competition Saturday, December 3, 5 A Show that every positive integer is a sum of one or more numbers of the form r 3 s, where r and s are nonnegative integers