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let $x_0$ = 2, $y_0 = 1$; $x_{n+1} = \frac{x_n + y_n}{2}$, $y_{n+1} = \frac{2}{x_{n+1}}, n = 0, 1, 2...$.

Need to show the sequences converge to $\sqrt 2$. I wish I could add more in terms of my attempted solution but I have really gotten anywhere; I haven't worked on problems like this in a while, and after spending time reviewing and attempting I haven't made much progress! Hints/solution would be appreciated.

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    But how to calculate $y_n$? – 2017-01-16
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    @SimpleArt that was dopey of me. I've corrected. – 2017-01-16
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    You sure the problem is written correctly? I'm not seeing anything that converges to the square root of 2. – 2017-01-16
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    I agree, I do not see any convergence to $\sqrt2$... – 2017-01-16
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    Which series needs to converge to $\sqrt{2}$? – 2017-01-16
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    Oops, I think I meant to write "sequence". – 2017-01-16
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    @S.C.B. both sequences – 2017-01-16
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    (1)....You cannot have written the problem correctly for if $x_n\to \sqrt 2$ then $y_{n+1}=2/x_{n+1}\to 2/\sqrt 2=\sqrt 2.$ But then $x_{n+1}=1/2(x_n+y_n)\to 1/2(2\sqrt 2)\ne \sqrt 2$..... (2).Since $y_{n+1}$ is a function of $x_{n+1}$ alone, we can substitute its expression as a function of $x_{n+1}$ into the formula for $x_{n+2}$ to obtain $x_{n+2}$ as a function of $x_{n+1}$ alone. – 2017-01-16
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    @user254665 oh geeze I've messed up a lot. The problem was badly typesetted in my defense. It's supposed to be $x_{n+1} = \frac{x_n + y_n}{2}$ – 2017-01-16
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    $x_{n+2}=(x_{n+1}+y_{n+1})/2=(x_{n+1}+2/x_{n+1})/2.$ Use induction on $n$ to show $2$x_{n+1}$ converges momtonically to a positive limit $L\geq \sqrt 2$. But so does $x_{n+2}.$ So $L=(L+2/L)/2.$ – 2017-01-16

2 Answers 2

1

We have directly the following:

$$x_{n+1}=\frac{x_n+\frac2{x_n}}2$$

So $y_n$ is really irrelevant. Now, if $L=\lim\limits_{n\to\infty}x_n$, then

$$L=\frac{L+\frac2L}2\implies L\stackrel?=\pm\sqrt2$$

Applying AM-GM inequality, we see that

$$x_{n+1}>\sqrt{x_n\frac2{x_n}}=\sqrt2$$

Thus, if $L$ exists, then it follows that it must be the square root of $2$.

To prove it exists, one may observe that if $x_n>\sqrt2$, then

$$\frac2{x_n}<\sqrt2$$

$$x_n+x_n>x_n+\sqrt2\implies x_n>\frac{x_n+\sqrt2}2$$

$$x_{n+1}=\frac{x_n+\frac2{x_n}}2<\frac{x_n+\sqrt2}2

Thus,

$$\sqrt2

So it converges, and to our desired value.

2

Note that you are attempting to calculate the limit of $x_{n}$ where $$x_{n}=\frac{1}{2} \left(x_{n-1}+\frac{2}{x_{n-1}} \right)$$

This is known as the Babylonian Method, and you can prove it yourself by proving that first, that is bounded below by $\sqrt{2}$, and then that $x_{n}$ is a decreasing sequence. This is relatively simple, and follows from $\text{AM-GM}$, so I leave to you.

This gives us the sequence must converge. Thus, note that if the limit is $a$ , $a$ satisfies $$2a=a+\frac{2}{a}$$ And you can use $x_{n}$ to compute $y_{n}$.

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    Oh, so your Babylonian method is basically what I did below? Hm, nifty. – 2017-01-16
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    @SimpleArt Yes, the Babylonian method is actually follows from Newton's method of approximating roots. – 2017-01-16