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I have been looking at this for hours and it isn't making anymore sense than it did in the first hour.

If $a$ und ${x_{0}}$ are positive real numbers and ${x_{k}}$ defined as follows, prove that ${x_{k}}$ is monotone decreasing and bounded, then calculate the limit.

${x_{k}} = \frac{1}{2}\left({x_{k-1}+\frac{a}{{x_{k-1}}}}\right)$

What I though I had to do was pick an ${x_{0}}$ and solve for ${x_{k}}$, so I picked ${x_{0}}$. Then I wanted to put the result back into the function to get ${x_{k+1}}$, which I still believe is what I'm supposed to be doing, but I don't understand what I am getting as a result. I get that I should prove it is decreasing, then that it is bounded, then address the limit, but the how is missing.

Martin Sleziak
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Lisa Ever
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2 Answers2

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If $x_{k-1}>\sqrt a$ then $\frac{a}{x_{k-1}} < x_{k-1}$ so $$x_k = \frac{1}{2}(x_{k-1}+\frac{a}{x_{k-1}}) < \frac{1}{2}(x_{k-1}+x_{k-1})=x_{k-1} $$ And $$x_k = \frac{1}{2}(x_{k-1}+\frac{a}{x_{k-1}}) \geq \frac{1}{2}2\sqrt a=\sqrt a $$ Therefore $x_k$ is always in between $x_{k-1}$ and $\sqrt a$, which means monotonic decrease. (And therefore $\frac{a}{x_{k-1}} < x_{k-1}$ always holds.)

Now even if you picked $x_{0}<\sqrt a$ then $$x_1 = \frac{1}{2}(x_{0}+\frac{a}{x_{0}}) \geq \frac{1}{2}2\sqrt a=\sqrt a $$ Therefore $x_k$ goes back to the above case from $k\geq1$.

Kay K.
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1

Hint:

If a sequence is recursively defined with a function $f$ defined on an interval $I$, it is monotonically increasing (resp. decreasing) if

  • $f(I)\subset I$;
  • on the interval $I$, the graph of $f$ is above (resp. below) the first bissectrix $y=x$, i.e. if $f(x)\ge x$ (resp. $f(x)\le x$) on $I$.

If furthermore, $f$ is continuous on $I$, the limit is a fixed point of $f$ on $I$.

Bernard
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