1873_solutions

# In this subsection to show that c n to prove that 1 n

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Unformatted text preview: n 1 xn  1  1  1 1 n1 j2 1 2 1 n j n n  11 j2 1 2 1 n1 j1 n1 1  x n 1 . j! d. Deduce that the sequence x n converges to a number between 2 and 3. Have you seen this number before? The student is being asked informally whether he/she recognizes that this limit is the number e. The number e will be seen in Chapter 10. 147 6. This exercise concerns the sequence x n defined by the fact that x 1  1 and that, for each n x n 1  a. 5 4x n 1 we have 2. Use Scientific Notebook to work out the first twenty terms in the sequence x n . b. Prove that 1 x n  2 for every n. We use mathematical induction. Since x 1  1, the assertion p 1 is true. Now suppose that n is any positive integer for which the assertion p n is true. We see that 1 5 2 41 5 4x n 2 5 42 2 2 and so the assertion p n1 is true. We deduce from mathematical induction that the assertion p n is true for every positive integer n. c. Prove that the sequence x n is strictly increasing. We use mathematical induction. For each positive integer n we take p n to be the assertion that x n  x n1 . Since x1  1  5 2  x2 we conclude that the assertion p 1 is true. Now suppose that n is any positive integer for which the assertion p n happens to be true. Then x n 2  5 4x n1 2 5 4x n 2  x n 1 and so the assertion p n1 must be true. We deduce from mathematical induction that the assertion p n is true for every positive integer n. d. Prove that the sequence has a limit x that satisfies the equation x 5 4x  2 on the interval Notebook to make a 2D plot of the expression equation x 5 4x  2  0 x5 4x  2  0. Ask Scientific 2, 2 and to solve the x 1, 2 numerically. Compare the answer obtained here with the results that you obtained in part a. 7. a. Given that for every number x  0, prove that f x x  3. fx  x  9 2x 2 3 for each n and that the equation f x  3 holds if and only if Solution: The desired result follows at once from the fact that whenever x  0 we have fx  x  9  2 2x b. Given that x 1  4 and, for each n prove that the sequence x n x 3 2x 2  3. 1, we have x n 1  x n  9 , 2x n 2 is decreasing and that the sequence converges to the number 3. Solution: Since x n1  f x n for each n and since f x  3 for every number x 3 we see at once that x n  3 for every n. To see that x n is decreasing we observe that if n is any positive integer then x2 9 xn  9 n  0. x n x n 1  x n 2x n 2 2 Since the sequence x n is a decreasing sequence in the interval 3, Ý we know that x n is convergent. If we write the limit of this sequence as x then it follows from the relationship 148 x n 1  x n  9 2x n 2 that x x  9 2x 2 from which we deduce that x  3. 8. This exercise is a study of the sequence x n for which x 1  0 and 1 x n 1  2  xn for every positive integer n. We note that this sequence is bounded below by 0 and above by 1/2. a. Supply the definition 1 2x to Scientific Notebook. Then open your Compute menu, click on Calculus, and choose to iterate the function f...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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