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induction

Course: FS 24, Fall 2009
School: Harvard
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Seminar Freshman 24i: Mathematical Problem Solving Some induction problems 1. It can be shown1 that every planar n-gon (n &gt; 3) P has an &quot;interior diagonal&quot; -- that is, two nonconsecutive vertices V, V such that the line segment joining V, V is contained in the interior of P . Use this to prove that the interior angles of P total (n - 2)180 (a.k.a. (n - 2) radians). [Which version of...

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Seminar Freshman 24i: Mathematical Problem Solving Some induction problems 1. It can be shown1 that every planar n-gon (n > 3) P has an "interior diagonal" -- that is, two nonconsecutive vertices V, V such that the line segment joining V, V is contained in the interior of P . Use this to prove that the interior angles of P total (n - 2)180 (a.k.a. (n - 2) radians). [Which version of induction is natural to use here?] 2. Recall that n k is the binomial coefficient (a.k.a. combinatorial coefficient) defined by n k = n! ; k! (n - k)! if k < 0 or k > n we set n k = 0. n k i) Given k 0 and n k, what is k + k+1 + k+2 + + k k k ii) [somewhat trickier] Given k 2 and n k, what is 1 k k ? + 1 k+1 k + 1 k+2 k + + 1 n k ? 3. For n > 0 and d 0, how many monomials of total degree d are there in n variables? For example, when d = n = 3 the number is 10: using variables x, y, z we find the cubic monomials x3 , x2 y, x2 z, xy 2 , xyz, xz 2 , y 3 , y 2 z, yz 2 , z 3 . (In particular the answer is not simply n : that's the number of degree-d monomials in which no d variable appears to power greater than 1.) 4. Give a formula for (cos x) (cos(2x)) (cos(4x)) (cos(8x)) (cos(2n x)). [Especially if you're seeing this for the first time, you might try to use the same trick to evaluate other products such as 89 cos(n ) = cos(1 ) cos(2 ) cos(3 ) cos(89 ), or to construct a problem along the same n=1 lines involving f (x)f (3x)f (9x)f (27x) f (3n x) for some function f .] 5. [An IMO problem, but it's from the Easiest IMO Ever, and we weren't told there that this was an induction problem . . . ] Find the integer solution (x, y) of (x2 + xy - y 2 )2 = 1 that has the largest value of x2 + y 2 subject to the conditions 0 x 1981, y 0 1981. (Follow-up: what can you say about the Diophantine equation2 (x2 + 4xy - y 2 )2 = 1?) 6. [Thanks to Sonal Jain for suggesting this one] For a S set of n (distinct) positive numbers, let (S) = { tT t | T S}; that is, (S) is the set of positive numbers that can be written as the sum of some (possibly empty) subset T S. Given n, how small can the cardinality #((S)) be? For example, if n = 1 or n = 2 then all 2n sums are distinct, and for n = 3 there can be at most one coincidence among the 23 sub-sums (if the largest element of S is the sum of the other two); so the minimal cardinality is 2, 4, 7 for n = 1, 2, 3 respectively. Let v be the left-most vertex (or one of them if there's a choice), and v , v its neighbors along the boundary of P . If v v is an interior diagonal, we are done. Else there is an interior diagonal vw for some other vertex w in the triangle forme by v, v , v ; for instance, we may choose for w the vertex in that triangle, other than v, v , v , that is closest to v. Thanks to Zach, our resident computational geometer, for finding this construction. Where did we use n > 3? With some more care we can even use this construction to prove that P has an "interior", that is, the fact (which I relegated in class to an application of the Jordan curve theorem) that P splits the plane into exactly two connected regions, an "interior" and an "exterior" of P . 2 That is, an equation to be solved in integers; Diophantus originally worked with rational numbers, but that can always be encoded into integer solutions as well by replacing an equation in rational numbers r1 , r2 , . . . , rn by a homogeneous equation in integers x0 , x1 , x2 , . . . , xn where ri = xi /x0 for each i = 1, 2, . . . , n. 1
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