Solutions to Two Interesting Problems

Transcription

1 Solutions to Two Interesting Problems Save the Lemming On each square of an n n chessboard is an arrow pointing to one of its eight neighbors (or off the board, if it s an edge square). However, arrows in neighboring squares (diagonal neighbors included) ma not differ in direction b more than 45 degrees. A lemming begins in [a] center square, following the arrows from square to square. How large must n be before it s possible for the lemming to sta on the board? [Winkler] SOLUTION 1. The net couple of pages prepare for the solution b introducing a simple set of definitions and lemmata. The solution itself is then quite short. We will stud the path separatel from the arrows it encounters. For eample, the illustration at right shows a closed path whose directions are not alwas given b the arrows. This path, shown b the gra squares, follows the sequence a5 b5 c6 d6 e5 f4 e3 d2 c2 c3 b4 a a b c d e f g We define the total circulation of a path to be the net change in directions of the arrows that are encountered along it. (Bear in mind that the circulation also depends, of course, on the field of arrows.) When going from one square to the net, the arrow s direction either stas the same (for a change of zero), twists 45 degrees to the right (for a change of -1), or twists 45 degrees to the left (for a change of +1). The shortest possible non-trivial paths are triangles, such as b4 a5 b5 b4. The arrows encountered on this triangular path occur in the sequence,,, and. The changes of direction are -1, +1, and 0. Its circulation therefore is = 0. We can combine paths. Because all we will care about is the circulation, and since that s defined in terms of changes along edges, a simple, rigorous wa to define path combination is to consider a path as an unordered, weighted collection of oriented edges: we will call this a chain. It doesn t matter what direction we assign to each edge on the board, but we need to make a consistent distinction between its two possible directions. One wa is to orient all edges so that motions to the North, Northeast, East, and Southeast are considered positive and motions in the other four cardinal directions are negative. In this sense, for eample, the illustrated path, thought of as a chain, can be Ma 02, 2008 W:\Haverford\Problems\2007-8\Spring\Miscellan.doc

2 written the sum of edges a5 b5, b5 c6, c6 d6, d6 e5, e5 f4, and c2 c3 minus the edges e3 f4, d2 e3, c2 d2, b4 c3, and a5 b4. Abstractl, chains are functions from the set of edges on the board into the integers. We combine two chains b adding them (as functions). For eample, the sum of the triangle b4 a5 b5 b4 and the triangle b4 b5 c5 b4 is equivalent to the chain of b4 a5 b5 c5 b4: the edges b5 b4 (in the first triangle) and b4 b5 (in the second triangle) cancel out. (This summation operation makes the set of chains into a commutative group.) Here is a list of obvious statements: ou are invited to make rigorous proofs of them. (1) The arrow at the end of a path differs from the arrow at the beginning b the total circulation modulo eight. (2) The circulation of a closed path does not depend on what square we start it at. (3) The circulation of a closed path must be a multiple of eight. (4) The circulation of a chain is well-defined. (5) The circulation of a sum of chains is the sum of the circulations of the chains. (That is, circulation is a homomorphism from the group of chains into the group of integers modulo eight.) (6) Ever closed path, as a chain, is the sum of triangles. (7) The circulation of an triangle is zero. Statement (4) is true because the circulation depends onl on changes in arrow directions encountered along individual edges. Because summing the changes is a commutative operation, the sequence of edges doesn t matter. Statement (6) is geometricall obvious. You could prove it b contradiction: consider the smallest path (in terms of number of edges) that cannot be epressed as a sum of triangles. Subtract a well-chosen triangle from it to obtain a et smaller path: b definition, it is a sum of triangles. Adding back the original triangle epresses the original path as a sum of triangles. Statement (7) is where we use the assumption that arrows in neighboring squares cannot differ b more than 45 degrees. That assumption had to show up somewhere! That ends the preparation. Statements (6) and (7) impl the total circulation of an closed path is zero. However, if the lemming is to sta on the board, it must follow a simple (that is, non-self-crossing) closed path whose circulation therefore is nonzero. This contradiction seals the poor lemming s fate. Page 2/8

3 SOLUTION 2. Winkler gives this solution: assume the lemming can sta on the board. Rotate all arrows b 45 degrees. The conditions of the problem still appl. If ou did the rotation in the correct direction, the lemming will be steered into the interior of the circular path it used to follow, creating a smaller path that keeps it on the board. However, this process cannot be carried out ad infinitum, because as we established in our discussions arbitraril small lemming circuits do not eist. Again we obtain a fatal contradiction. Eponential tower Let =. Make sense of this epression; find the (positive) real values for which it is well-defined; find the values for which it is differentiable; and find its derivative. SOLUTION. The value of this tower of eponents depends on where the implicit parentheses are. Two natural choices are ( ) ( ) ( ) = and = ( (( ) ) ). The latter is uninteresting, because it makes equal to the limit of the sequence,, ( 2 ) =, ( 2 3 ) =,, n,. This has a limit for positive onl when = 1, where obviousl = 1. Therefore we choose to analze the first definition, which is the limit of the sequence,,,. Generall, the n+1 st term, a n+1, is obtained from the n th term, a n, via a = If (a n ) has a limit, then evidentl We solve this for in terms of : a n n+1 ; equivalentl, setting b n = ln(a n ), b n+1 = ln() ep(b n ). = ; equivalentl, ln() = ln(). ln() = ln() / ; Page 3/8

4 = ep(ln() / ) = 1 / = f(), sa. This function f is defined for all positive and, as the composition of differentiable functions ep, ln, and division, is differentiable. Let s analze it. An eas wa to find its derivative is to compute the derivative of its logarithm, because for general positive differentiable functions f the derivative of ln(f()) is f '() / f(). Therefore For later reference, note that (*) f '() / f() = d/d ( ln() / ) = (1 ln()) / 2. f '() = 1 / (1 ln()) / 2 = 1/ 2 (1 ln()). The critical points of f occur where the derivative is zero, as approaches 0, and as approaches infinit. To find these limits we eploit the continuit of ln and ep: f '() = 0 is equivalent to 1 ln() = 0, with unique solution = e; lim 0 f ( ) = ep( lim ln( f ( )) ) = ep( lim ln( ) / ) = lim ep( ) = 0; and 0 0 lim f ( ) = ep( lim ln( ) / ) = ep(0) = 1. Furthermore, the limiting value of f '() as approaches zero equals lim 0 f ' ( ) lim = ( ) 2 0 f ( ) 1 ln( ) /. Taking logarithms, we see that ln(f '()) is the sum of three terms: ln(f()) = ln()/; ln(1 ln()), which behaves like ln(-ln()) for small ; and -2ln(). The first term dominates the other two, impling the slope of f approaches zero at the origin. We conclude that f has local minima of 0 and 1 at the limits 0 and, respectivel, and attains its onl local maimum (which is therefore a global maimum) at = e, where it equals e 1/e This information lets us draw a graph that is accurate enough to infer properties of f. Page 4/8

5 The inverse of f, (), should be the infinite tower of eponentials. But f is not invertible. What s going on? We realized quickl that the tower of eponentials does not converge for all values of. To understand its convergence, we analzed the tower as a dnamical sstem. For an given value of, let a = ln(). Fi this value and consider the function g(t) = ep(at), t > 0. Beginning at t = 1, repeated applications of g give the sequence 1, g(1) = ep(a) =, g() = g 2 (1) = ep(a ) =, g( ) = g 3 (1) =, etc. In other words, iterating g for k times gives a tower of k eponentials based on. The infinite tower eists, b definition, if and onl if the sequence (1, g(1), g 2 (1),, g k (1), ) converges. We developed a graphical method to visualize this procedure. The sequence of iterates defines a sequence of ordered pairs (1, g(1)), (g 2 (1), g(1)), (g 2 (1), g 3 (1)), (g 2k (1), g 2k-1 (1)), (g 2k (1), g 2k+1 (1)), Notice how we have reversed the order in ever other pair. Let h = g -1. The graph of h, as alwas, is obtained b flipping the graph of g across the diagonal. Fortunatel, h is also a function, because We rewrite the sequence of ordered pairs as h() = ln() / a. Page 5/8

6 (1, g(1)), (g 2 (1), h(g 2 (1)), (g 2 (1), g(g 2 (1))), Schematicall, this follows a zig-zag sequence: begin at (1, 0). Move verticall until ou hit the graph of g at (1, g(1)). Move horizontall until ou hit the graph of h at (g 2 (1), h(g 2 (1)). Repeat this procedure ad infinitum. Here s a picture with = The blue (mostl upper) line is the graph of g and the red line is the 0.2 graph of h. The solid light line etending up and to the right from 0.1 (0.4, 0.4) is the main diagonal: the red line is obtained b reflecting the 0.0 blue line around it. The dotted line shows how the dnamical sstem works. Here, it proceeds from (1, ) = (1, 0.25) leftwards to ( 2, ) = (0.707, 0.25), then up, etc. Evidentl it converges at a point where all three solid lines meet: = t = ep(a t) = ln(t) / a. This common value is the value of the tower of eponentials for. One thing that made this problem so interesting is that when is sufficientl small, the dnamical sstem does not converge but the geometric construction does! This figure shows the situation when = The geometric construction converges to a point near (0.7495, ). This means the dnamical sstem alternates between two accumulation points without approaching a limit. Indeed, = and = Page 6/8

7 Geometricall, it s now obvious that the sstem bifurcates from a domain of convergence into this alternating regime when the graphs of g and h are tangent where the cross. We can calculate when this occurs, because we 1.0 know (1) g'(t) = h'(t) = -1 and (2) t = t Computing the derivative of h, then, we obtain equations 1/(t ln()) = h'(t) = -1, t = t These impl ln(t) = ln( t ) = t ln() = -1, so t = 1/e; ln() = -1/t = -e, so = e -e The analsis for > 1 is simpler: the dnamical sstem converges (and thereb provides a rigorousl defined value for the eponential tower) when e 1/e , and otherwise diverges. The bifurcation occurs when (1) g'(t) = h'(t) = +1 and (2) t = t, impling (using the same method of solution as before) that t = e and = e 1/e. The graph below at the left shows the dnamical sstem at the bifurcation point = At the right, ou can see how the sstem slips out of the finite region (for = 1.48), because the graphs of g and h no longer meet. The tower of eponentials diverges for this value of. Page 7/8

8 Now that we are satisfied the tower converges for all in the interval [e -e, e 1/e ] and does not converge for an other, and having alread established that its inverse is f() = 1/, we finall can answer the original question about its derivative, '()! Appling the Inverse Function Theorem, and using the epression (*) for f '() obtained several pages back, we get (**) d/d () = 1 / f '(()) = 2 1/ /(1 ln()). (If ou prefer to see our formulas for derivatives in terms of itself, just make the substitution =. It s ugl, so we won t tr to print it here.) For eample, just to show we have obtained a practical result, consider the case = 2 = 2 1/2 : 2 1/2 < e 1/e implies (2 1/2 ) = 2, so '(2 1/2 ) = 2 2 1/2 /(1 ln(2)) = 2 3/2 /(1 ln(2)). At the endpoints of convergence, (e 1/e ) = e, so '(e 1/e ) = (something finite)/(1 e/e) is undefined and (e -e ) = 1/e, so '(e -e ) = (1/e) 2 e / (1 + 1) = e e 2 /2. Consequentl, the tower of eponentials is defined on the closed interval [e -e, e 1/e ], is the inverse of 1/ there, and is differentiable on the half-open interval [e -e, e 1/e ) with derivative given b the formula (**) above. Page 8/8