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In #5, I'm having trouble figuring out how to prove that sequence <math>Z</math> is bounded in order to use Bolzano-Weierstrass Theorem and Theorem 3.4.9 to prove the necessary and sufficient conditions. I know that since both sequences <math>X</math> and <math>Y</math> are convergent that they are bounded, but I can't quite figure out how to use this information to prove that <math>Z</math> is bounded. | In #5, I'm having trouble figuring out how to prove that sequence <math>Z</math> is bounded in order to use Bolzano-Weierstrass Theorem and Theorem 3.4.9 to prove the necessary and sufficient conditions. I know that since both sequences <math>X</math> and <math>Y</math> are convergent that they are bounded, but I can't quite figure out how to use this information to prove that <math>Z</math> is bounded. | ||
| − | --You can try to approach the problem in a different way. Lets say x and y converge to L. Show that for some n>a, abs(x-L)<e and n>b, abs(y-L)<e. Then show that there exists some c for which, n>c, abs(z-L)<e! Hope n>a | + | -- |
| + | Siddharth Tekriwal: | ||
| + | |||
| + | You can try to approach the problem in a different way. Lets say x and y converge to L. Show that for some n>a, abs(x-L)<e and n>b, abs(y-L)<e. Then show that there exists some c for which, n>c, abs(z-L)<e! Hope n>a | ||
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In #5, I'm having trouble figuring out how to prove that sequence $ Z $ is bounded in order to use Bolzano-Weierstrass Theorem and Theorem 3.4.9 to prove the necessary and sufficient conditions. I know that since both sequences $ X $ and $ Y $ are convergent that they are bounded, but I can't quite figure out how to use this information to prove that $ Z $ is bounded.
-- Siddharth Tekriwal:
You can try to approach the problem in a different way. Lets say x and y converge to L. Show that for some n>a, abs(x-L)<e and n>b, abs(y-L)<e. Then show that there exists some c for which, n>c, abs(z-L)<e! Hope n>a
In #3, what will be a good starting point? I am having difficulty proceeding through the problem.
--Rrichmo 20:15, 10 March 2010 (UTC)
What I did was write out the sequence (Xn) to notice that (Xn+1) (the next element in the sequence) is equal to 1 + 1/Xn (this was not intuitive to me). Knowing that both (Xn) and (Xn+1) equal the same limit you can set them equal to each other like in example 3.4.3, then solve.
More detail about this solution method can also be found at http://www.mathacademy.com/pr/prime/articles/fibonac/index.asp
M.N.
