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a) What happens as you take repeated square roots, starting with 8?
b) What does the answer approach as you take more and more square roots?
c) Would the answer be the same if you started with any number? Explain your answer.
Plz give detailed answers to this ques in parts and explain them all tyvm guys and plz give the answer asap

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9  
How many times have you computed repeatedly the square roots, starting from $8$? I suggest you should do it at least 4-5 times. You can find some calculator program on the net if your computer misses it. Please, experience it yourself, ASAP! –  Berci Feb 21 '13 at 11:35
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Would you like fries with that sir? –  Thomas Feb 21 '13 at 11:35
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Plz? ques? tyvm? We prefer people to write like grownups on this website. –  Gerry Myerson Feb 21 '13 at 11:41
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@Abhishek: The intention of this site is to help people with maths, rather than answer their homework questions for them. Please pay attention to Berci's comment. –  Tara B Feb 21 '13 at 11:49
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Please don't make your post unintelligible after receiving answers. –  Martin Feb 24 '13 at 2:16

2 Answers 2

up vote 5 down vote accepted

When you take the $\sqrt{8}$ you get $2\cdot \sqrt{2}$, taking again the square root gives you $\sqrt{2} \cdot \sqrt{\sqrt{2}}$.

Here are some numerical values for that: \begin{array}{c} 8\\ 2.82843 \\ 1.68179 \\ 1.29684 \\ 1.13879 \\ 1.06714 \\ 1.03302 \\ 1.01638 \\ 1.00816 \\ 1.00407 \\ 1.00203 \\ \end{array}

If you make that often enough, the sequence will converge to $1$.

The answer is the same if you start with any number $> 0$, for $0$ you always have $0$.

An easier way is maybe that for $x>4$ we know $$\sqrt{x}\leq \frac{x}{2},$$ And we know that for $x\leq \frac{1}{4}$ $$\sqrt{x} \geq 2\cdot x$$ So it is enough if we check the behaviour in $(\frac{1}{4},4)$, as the square root is monotone, this one works if we only check it for $\frac{1}{4}$ and for $4$

If you like it more formal, here is another solution. The square root is a contraction, that means if you take any $x_0$ and let $x_{i+1}=\sqrt{x_i}$, than $$|x_{i+1}-x_i| < |x_i-x_{i-1}|,$$

to say it in words, the difference between those values, always get smaller. There is a theorem called Banach Fixpoint-Theorem, which says, when a function is a contraction on a closed subset $M$ of $\mathbb{R}$, and $f:M \rightarrow M$ then there is exactly one point which satisfies $f(x)=x$ as you know that $\sqrt{1}=1$ it will be that one.

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2  
In fact, if you work on the complex plane, it works for any nonzero complex number (even negative reals). –  Thomas Feb 21 '13 at 11:40
    
This answers the questions, except for the bit about "explain". –  Gerry Myerson Feb 21 '13 at 11:42
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This is a sequence, not a series. –  mrf Feb 21 '13 at 11:46
    
Yes, but 1) the inequality is just an unsupported assertion, and 2) Banach seems like overkill. –  Gerry Myerson Feb 21 '13 at 11:50
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+1 just for the fact that you took some of your time to answer a question formulated like that. –  Djaian Feb 21 '13 at 12:15

I did it asap given that I've just seen your question...

Here is an informal approach, first. Your sequence is defined recursively. A picture is worth a thousand words.

Plot $f(x)=\sqrt{x}$ on $[0,+\infty)$, together with $g(x)=x$.

You'll see that there are two intersections at $0$ and $1$, of course. On $(1,+\infty)$, $f$ is below $g$. On $(0,1)$, $f$ is above $g$.

Start with any $x_0$ in $(1,+\infty)$, like $8$.

Then move up to the graph of $f$. You're at the point $(x_0,\sqrt{x_0})$.

Move left to the graph of $g$. This is now the point $(\sqrt{x_0},\sqrt{x_0})$.

Move down to the graph of $f$. Here is $(\sqrt{x_0},\sqrt{\sqrt{x_0}})$.

Mover left to the graph of $g$....

Continue these steps repeatedly.

You'll see that the first coordinate, which is your sequence $$ x_n=f^n(x_0)=\sqrt{\sqrt{\ldots\sqrt{x_0}}} $$ is decreasing and converges to $1$.

Note that if you start with $x_0\in(0,1)$, then $x_n$ is increasing and converges to $1$.

If you start with $x_0=0$ or $1$, the sequence is constant.

Of course, once you get the idea, the formal proof is very easy.

Prove that if $x_0\in(1,+\infty)$, $x_n=f(x_{n-1})$ remains in $(1,+\infty)$ for all $n$, is decreasing, and bounded below by $1$. This follows from an easy study of $f$ and some induction.

Then it must converge to a fixed point of $f$ which is $\geq 1$. So this has to be $1$.

Then treat the case $x_0\in(0,1)$ in a simmmilar fashion, except that now the sequence is increasing and bounded above by $1$.

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ty guys i just needed to understand the question more properly really appreciate it! Now i can answer my questions in assignment more easily :) –  Abhishek Sehgal Feb 24 '13 at 1:50

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