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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! | 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! | ||
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If you want to show that Z is bounded you can use that X,Y are bounded by Theorem 3.2.2 and let X<a, and Y<b then let c=max(a,b) then Z<c, because if z is in Z then z is in X if n is odd or z is in Y if n is even, thus z<=a, or z<=b for any z in Z. Then by definition Z is bounded | If you want to show that Z is bounded you can use that X,Y are bounded by Theorem 3.2.2 and let X<a, and Y<b then let c=max(a,b) then Z<c, because if z is in Z then z is in X if n is odd or z is in Y if n is even, thus z<=a, or z<=b for any z in Z. Then by definition Z is bounded | ||
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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! ---
If you want to show that Z is bounded you can use that X,Y are bounded by Theorem 3.2.2 and let X<a, and Y<b then let c=max(a,b) then Z<c, because if z is in Z then z is in X if n is odd or z is in Y if n is even, thus z<=a, or z<=b for any z in Z. Then by definition Z is bounded
M. Niekamp
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.
-- Siddharth Tekriwal:
you can use the fibonacci equation. f(n+2) = f(n+1) + f(n) Therefore (fn+2)/(fn+1) = 1 + (fn)/(fn+1). You can then try to prove that both terms are not unbounded and then apply the quotient theorem.
