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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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| + | How can you calculate the limits of the sequences in #7? Is there some trick I'm not aware of? | ||
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| + | Hi. These limits are all consequences of the Euler limit: | ||
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| + | <math> \lim \left( 1+\frac{1}{n}\right)^{n} = e </math> | ||
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| + | (a) This sequence is a subsequence of <math> x_{n_{k}} </math> of <math> x_{n} </math> for | ||
| + | <math>n_{k}=k^2</math>. Indeed, we just keep the terms whose numbers are perfect squares. | ||
| + | So, the limit is <math> e \, </math>: | ||
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| + | <math> \lim \left( 1+\frac{1}{n^2}\right)^{n^2} = e </math> | ||
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| + | (b) This can be easily derived from the part (a): | ||
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| + | <math> \lim \left( 1+\frac{1}{n^2}\right)^{2n^2} = \lim \left(\left( 1+\frac{1}{n^2}\right)^{n^2}\right)^2 = </math> | ||
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| + | <math> \left( \lim \left( 1+\frac{1}{n^2}\right)^{n^2}\right)^2 = e^2 </math> | ||
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| + | Can somebody do (c) and (d) | ||
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| + | Thanks, Prof. Alekseenko --[[User:Aaleksee|Aaleksee]] 15:58, 11 March 2010 (UTC) | ||
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| + | c)<math> \lim \left( 1+\frac{1}{2n}\right)^{n}</math> | ||
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| + | This sequence is a subsequence of <math> x_{n_{k}} </math> of <math> x_{n} </math> for | ||
| + | <math>n_{k}=2k</math>. Indeed, we just keep the terms which are even. | ||
| + | So, the limit is <math> e \, </math>: | ||
| + | |||
| + | <math> \lim \left( 1+\frac{1}{2n}\right)^{n} = x_{n}^{0.5} = e^{0.5} </math> | ||
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| + | d)<math> \lim \left( 1+\frac{2}{n}\right)^{n}</math> | ||
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| + | This sequence is a subsequence of <math> x_{n_{k}} </math> of <math> x_{n} </math> for | ||
| + | <math>n_{k}=k/2</math>. | ||
| + | So, the limit is <math> e \, </math>: | ||
| + | |||
| + | <math> \lim \left(\left( 1+\frac{1}{n/2}\right)^{n/2}\right)^2 = x_{n}^{2} = e^{2} </math> | ||
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| + | Yuvraj Singh | ||
Latest revision as of 19:51, 20 March 2010
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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.
How can you calculate the limits of the sequences in #7? Is there some trick I'm not aware of?
Hi. These limits are all consequences of the Euler limit:
$ \lim \left( 1+\frac{1}{n}\right)^{n} = e $
(a) This sequence is a subsequence of $ x_{n_{k}} $ of $ x_{n} $ for $ n_{k}=k^2 $. Indeed, we just keep the terms whose numbers are perfect squares. So, the limit is $ e \, $:
$ \lim \left( 1+\frac{1}{n^2}\right)^{n^2} = e $
(b) This can be easily derived from the part (a):
$ \lim \left( 1+\frac{1}{n^2}\right)^{2n^2} = \lim \left(\left( 1+\frac{1}{n^2}\right)^{n^2}\right)^2 = $
$ \left( \lim \left( 1+\frac{1}{n^2}\right)^{n^2}\right)^2 = e^2 $
Can somebody do (c) and (d)
Thanks, Prof. Alekseenko --Aaleksee 15:58, 11 March 2010 (UTC)
c)$ \lim \left( 1+\frac{1}{2n}\right)^{n} $
This sequence is a subsequence of $ x_{n_{k}} $ of $ x_{n} $ for $ n_{k}=2k $. Indeed, we just keep the terms which are even. So, the limit is $ e \, $:
$ \lim \left( 1+\frac{1}{2n}\right)^{n} = x_{n}^{0.5} = e^{0.5} $
d)$ \lim \left( 1+\frac{2}{n}\right)^{n} $
This sequence is a subsequence of $ x_{n_{k}} $ of $ x_{n} $ for $ n_{k}=k/2 $. So, the limit is $ e \, $:
$ \lim \left(\left( 1+\frac{1}{n/2}\right)^{n/2}\right)^2 = x_{n}^{2} = e^{2} $
Yuvraj Singh
