Calculus:Theory And Applications by Kenneth Kuttler

By Kenneth Kuttler

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Here is how you might proceed. If not, ab = [a, b] q + r where 0 < r < [a, b] . Then verify r is a common multiple of a and b contradicting that [a, b] is the least common multiple. Hence r = 0. Therefore, [a, b] = ab/q for some q an integer. Since [a, b] is a common multiple of a and b, argue that q must divide both a and b. Now what is the largest such q? This would yield the smallest ab/q. You fill in the details. 5. Show that if {a, b, c} are three positive integers, they have a greatest common divisor which may be written as ax + by + cz for some integers x, y, z.

For all n ≥ 0. Prove that n whenever k ≥ 1 and k ≤ n, then n+1 = nk + k−1 . Are these numbers, k same as those obtained in Pascal’s triangle? Prove your assertion. n k the n n 18. The binomial theorem states (a + b) = k=0 nk an−k bk . Prove the binomial theorem by induction. Hint: You might try using the preceding problem. n n k=0 k 19. Show that for p ∈ (0, 1) , n−k kpk (1 − p) = np. 20. Using the binomial theorem prove that for all n ∈ N, 1 + n k Hint: Show first that = n·(n−1)···(n−k+1) . k!

C ✟ ✟ ✟ ✟ ✟✟ A ✟✟ b ✟ ✟✟ a ✟ ✟ c C ✟ ✟ b ✟ a ✟✟ B A✟ c B Denote by a, a , b, b , c and c the sides indicated in the picture. Then by the law of cosines, a2 = b2 + c2 − 2bc cos A = 2b2 − 2b2 cos A and so a/b = 2 (1 − cos A). Similar reasoning shows a /b = 2 (1 − cos A) and so a/b = a /b . Similarly, a/c = a /c . By assumption c/b = 1 = c /b . Such triangles in which two sides are equal are called isoceles. 5 The Circular Arc Subtended By An Angle How can angles be measured? This will be done by considering arcs on a circle.

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