What is the fastest way to compute the square-root of a real, positive number x to a desired number of correct digits?  For example, in number theory, $\sqrt{x}$ for nonnegative integers $x$ can be approximated to desired accuracy as a convergent of the continued fraction of $\sqrt{x}$ if this is known. In calculus,  $\sqrt{t}$ can be found using the bisection method. This method is slow. Now, $\sqrt{t}$ can be computed faster by applying Newtons' method to the polynomial $x^2-t$ for any initial guess $k$ not equal to zero. When I say "fastest" I mean "least amount of computations" :)  I also learned about Taylor polynomials; maybe using a Taylor polynomial about some $x$ close to to where $\sqrt{x}$ is known can be a fast method?
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What is the fastest algorithm for finding the square root of a number?  I created one that can find the square root of "$987654321$" to $16$ decimal places in just $20$ iterations   I've now tried Newton's method as well as my own method (Newtons code as seen below)  What is the fastest known algorithm for taking the second root of a number?  My code for Newton's Method (*Edit: there was an error in my code, it is fixed in the comments below):        a=2   //2nd root     b=97654321   //base     n=1   //initial guess     c=0   //current iteration (this is a changing variable)     r=500000   //total number of iterations to run     while (c&lt;r)      {         m = n-(((n^a)-b)/(a*b))  //Newton's algorithm         n=m         c++;         trace(m + "  &lt;--guess   ...   iteration--&gt;  " + c)     }
71721
Do you a digit efficient way to approximate $\pi$? I mean representing many digits of $\pi$ using only a few numeric digits and some sort of equation. Maybe mathematical operations also count as penalty.  For example the well known $\frac{355}{113}$ is an approximation, but it gives only 7 correct digits by using 6 digits (113355) in the approximation itself. Can you make a better digit ratio?  EDIT: to clarify the "game" let's assume that each mathematical operation (+, sqrt, power, ...) also counts as one digit. Otherwise one could of course make artifical infinitely nested structures of operations only. And preferably let's stick to basic arithmetics and powers/roots only.  EDIT: true. logarithm of imaginary numbers provides an easy way. let's not use complex numbers since that's what I had in mind. something you can present to non-mathematicians :)
19713
Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.     $$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
50304
Last year in Pre-Algebra we learned about square roots. I was taught then that $\sqrt{64}=8$ and $\sqrt{100}=10$, which I understood and accepted. I was also taught that $\pm\sqrt{64} = 8,-8$ because both of those numbers squared is 64, which I also get. But this year, with a new school and teacher in a different state, our teacher is telling us that: $\sqrt{64}=8,-8$ and $\pm\sqrt{64}$ also is $8,-8$. The way to get the positive root of something is: $+\sqrt{64}=8$ And these seem to contradict each other. I was always taught that a regular square root returned a positive number and only a positive number, but now my teacher is saying a regular square root gives two numbers, and considering the square root of a number $n$ is defined as $y^2=n$ I see where he is coming from. Upon researching this Wikipedia says:  For example, $4$ and $−4$ are square roots of $16$ because $4^{2} = (−4)^{2} = 16$  And Wolfram MathWorld says:  Note that any positive real number has two square roots, one positive and one negative. For example, the square roots of $9$ are $-3$ and $+3$  But on the other side, Wolfram Alpha, when given &quot;The square root of 9&quot; gives only 3.  So, which is right? Is $\sqrt{64}$ considered $8$? or is it $8,-8$?
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Suppose that we don't know logarithm, then how we would able to calculate $\sqrt x$, where $x$ is a real number? More generally, is there any algorithm to calculate $\sqrt [ n ]{ x } $ without using logarithm? More simple techniques would be nice.  Here is a simple technique used to approximate square roots by :  For example: $\sqrt {78}\approx 8\frac { 14 }{ 17 } $ , where $8$ is the nearest integer root of $78$, $14 = 78 - 8^2$, $17 = 2 \times 8 + 1$.  if $n=2^k$ we can use the method above.   For example, for $k=2$ Lets calculate $\sqrt [ 4 ]{ 136 } $: $$\sqrt [ 4 ]{ 136 } =\sqrt { \sqrt { 136 }  } \approx \sqrt { 11\frac { 136-{ 11 }^{ 2 } }{ 11\times 2+1 }  } =\sqrt { 11\frac { 15 }{ 23 }  } \\ \sqrt { 11\frac { 15 }{ 23 }  } \approx 3\frac { 11\frac { 15 }{ 23 } -{ 3 }^{ 2 } }{ 3\times 2+1 } =\frac { 544 }{ 161 } =3.38\\$$ The exact result is$$ \sqrt [ 4 ]{ 136 } =3.4149\cdots$$ The method approximates well, but it is working for only $n=2^k$ as I know.
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I am aware that 1/0 is undefined for two reasons:   If you would have to give an answer to this it is infinity which is not a number but a concept; The limit of 1/x for $x \to 0$ is either positive or negative infinity and since it has two limits we cannot state what its exact value is   How can we use this for square roots? It has multiple answers so why do we pick the positive one? if $x^2 = 16 \implies x = \sqrt{16} $ or $x = -\sqrt{16}$ for respectively the positive and negative solution. This implies that the square root function has a single answer and we must negate its answer to obtain the second solution. I understand this is extremely practical but it feels like 'cheating' since we know that there are 2 solutions, but in the function we simply ignore one of them. Is this correct?
3243
For any $a &gt; 0$, I have to show the sequence $x_{n+1}$ $=$  $ \frac 12$($x_n+ $ $ \frac {a} {x_n}$)  converges to the square root of $a$ for any $x_1&gt;0$.  If I assume the limit exists ( denoted by $x$) then,  $x$ $=$  $ \frac 12$($x+ $ $ \frac {a} {x}$) can be solved to $x^2 = a$  How could I show that it does exist?
34010
$$T(n) = \sqrt{n} T \left(\sqrt n \right) + n$$  Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable.  The answer is $Θ (n \log \log n)$.  Can anyone arrive at the solution.
76777
I have no idea how to do this.  To find the minimal polynomial of say $\sqrt2 + \sqrt3$, we need to find the monic polynomial $p \in \mathbb Q$ (correct if I am wrong but monic polynomial is when the coefficient of the highest degree term is $1$) of the smallest possible degree such that $\sqrt2 + \sqrt3$ is a root of $p$.  If we let $u=\sqrt2 + \sqrt3$ then $u ^2 = 5+ 2 \sqrt6 \iff u^2 - 5 = 2 \sqrt6 $, then $(u^2 - 5)^2=24 \iff u^4 -10u^2 +1=0$  All I did was keep squaring until all of the irrational terms go away. But what next? Am I doing this correctly and what do we do next if I am?
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Where $i$ is the square root of negative one.  And is there a generalization of the $n$th root of $i$? Also how would this look graphically on the real number axis? Thanks
123424
A good approximation of $(1+x)^n$ is $1+xn$ when $|x|n &lt;&lt; 1$.  Does this approximation have a name?  Any leads on estimating the error of the approximation?
74365
Can the square root of a real number be negative?  Dealing with the questions of functions in eleventh class my maths teacher says that square root of a real number is always positive. How is it possible?
33636
Suggest algorithm for the numerical calculation $ \frac{1}{\sqrt{a}} \ a &gt; 0 $ without division, use Newton's method.  My idea is:  $$ \frac{1}{\sqrt{a}} = (\sqrt{a})^{-1} = a^{-\frac{1}{2}}$$ $$ x = a^{-\frac{1}{2}} \Rightarrow f(x)=x^2 - a^{-1}$$ And next step is calulate $ f'(x)=2x $, but after substitution I got: $$ x_{n+1} = \frac{x_n^2+a^{-1}}{2x_n}$$ And know I still have divion, I guess. Maybe you've better idea on this task?
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Consider the function $f(x)=\sqrt x$.  Pretty much every resource I can find gives its domain as $\{x\in \mathbb{R} \mid x \ge 0\}$ and its range as $\{y\in \mathbb{R} \mid y \ge 0\}$  I don't see why the domain needs to be restricted to positive numbers. Isn't the square root of a negative number defined? That is, the square root of a negative number is a complex number of the form $a + bi$.  So why isn't the domain of $f(x)=\sqrt x$ instead $\{x\in \mathbb{R}\}$ and its range $\{y\in \mathbb{C} \mid y \ge 0\}$?  What am I missing?
57633
Why do we use only square root approach to find a number is prime or not? why not cube root &amp; 4rth root?
109555
In class we are currently covering upper/lower bounds, big Oh and omega and the like. I am pretty good on the "typical" functions one would do, but at a complete loss at "general" statements. This ons function, for example:  $$\forall f\in\mathcal{F}: \lfloor \sqrt{\lfloor f(n)\rfloor }\rfloor \in O(\sqrt{f(n)}).$$  the domain F by the same is all natural numbers that return a real positive number when inputted. My first instinct with something like this is to remove the floors, the general strategy I though to solve this is remove the outer floor, then get rid of the square root, and then the inner floor. But I have no idea how to go about this or if this is even the right way to do it.  Help please? Thanks in advance.
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How to compute $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}$$ &amp; $$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}$$?  I understand that  $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}=(1^{1/2})(2^{1/4})(3^{1/8})\cdots$$  and  $$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}=(1+(2+(3+(\cdots))^{1/8})^{1/4})^{1/2} $$  How to Proceed further?