By searching I found few methods but all of them involve guessing which is not what I want.
I need to know how to calculate the square root using a formula or something. In other words how does the calculator calculate the square root?
By searching I found few methods but all of them involve guessing which is not what I want.
I need to know how to calculate the square root using a formula or something. In other words how does the calculator calculate the square root?
There's one algorithm which is not bad at all, and it's stable. It involves no guessing, but you do need a starting point. To illustrate the algorithm, suppose you want to find $\sqrt{2}$. Start with anything (the closer the better); for example start with $1$. Because $1 < \sqrt{2}$, then $2/1=2 > \sqrt{2}$. Average these to get $1.5$. $1.5$ is too large, which means that $2/1.5=4/3=1.3333\cdots$ will be smaller than $\sqrt{2}$. Average these two to obtain $\frac{1}{2}(3/2+4/3)=\frac{1}{2}\frac{17}{6}=\frac{17}{12}=1.416666\cdots$. This value is too small, and $2/(17/12)=\frac{24}{17}$ is too large. So average these to get $$ \frac{1}{2}\left(\frac{17}{12}+\frac{24}{17}\right)=\frac{577}{408}=1.4142156862\cdots $$ This is already accurate to 6 decimal places. Repeating this process leads quickly to a good approximation of $\sqrt{2}$. One more iteration gives $$ \frac{1}{2}\left(\frac{577}{408}+2\frac{408}{577}\right) =\frac{665857}{470832}=1.4142135623747\cdots, $$ which is accurate to about 12 digits. The starting guess of $1$ wasn't all that great, but it still didn't take many iterations to get a good value of $\sqrt{2}$.
One method for approximating roots is Newton's Method, which uses a tangent line approximation to approach the actual value of a root.
Since it is given that $f(x) \approx f(x_0) + (x - x_0)f'(x_0)$, consider the function $f(x) = x^2 - 2$. Clearly, this quadratic has root $\sqrt{2}$. The goal is to get closer and closer to the y-intercept of this function (which is $\sqrt{2}$). We do this by taking the tangent line approximation of a value on the curve $x^2 - 2$ (say, $1.5$) and finding where that line intercepts the y-axis. Then, we plug the intercept in and get a new (more accurate) intercept.
(source: lamar.edu)
Check Paul's Online Notes for more info, or Wikipedia.
But to answer your question definitively, no. There is no way to get the value of an irrational number like $\sqrt{2}$ through a simple formula; it nearly requires computation. :(
The easiest way to find $\sqrt[n]{a}$ for integer $n$ and $a>0$ efficiently is to use the Newton-Raphson approximation to invert the function $f : x \mapsto x^n - a$. But one must be careful with choosing the right starting point, so that the iteration will converge quadratically. Quadratic convergence means that at each step the error becomes approximately a constant times its square, which is equivalent to the error being proportional to $c^{2^k}$ after $k$ steps, for some $c \in (0,1)$
Let $x_0$ be such that $x_0 \in \sqrt[n]{a}[1,1+\frac{1}{4n})$
For each natural $k$ from $0$ to $\infty$:
Let $x_{k+1} = x_k - \dfrac{f(x_k)}{f'(x_k)} = x_k - \dfrac{{x_k}^n-a}{n{x_k}^{n-1}} = \dfrac{(n-1){x_k}^n-a}{n{x_k}^{n-1}}$
Then $( x_k : k \in \mathbb{N} )$ converges quadratically to $\sqrt[n]{a}$ uniformly for all $a>0$
General Case
For any real function $f$ such that $f(r) = 0$ and $f' \ne 0$ and $f''$ exists and $\left|\frac{f''}{2f'(r)}\right| \le m$ for some $m$:
Let $a = f'(r) \ne 0$
Then $f(r+d) = a d + g(d) d^2$ for any $d$ for some function $g$ such that:
$g(d) \in a [-m,m]$ for any $d$
Also $f'(r+d) = a + h(d) d$ for any $d$ for some function $h$ such that:
$h(d) \in a [-m,m]$ for any $d$
Let $( x_k : k \in \mathbb{N} )$ be such that:
$x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)}$ for any natural $k$
$|x_0-r| \le \frac{1}{6m}$
For each natural $k$ from $0$ to $\infty$:
$x_k = r + d_k$ for some $d_k$
$|d_k| \le |d_0| \le \frac{1}{6m}$ by invariance
$x_{k+1} = (r+d_k) - \dfrac{ad_k+g(d_k){d_k}^2}{a+h(d_k){d_k}} \in (r+d_k) - \dfrac{d_k+[-m,m]{d_k}^2}{1+[-m,m]{d_k}}$
Thus $d_{k+1} \in d_k - (d_k+[-m,m]{d_k}^2) (1-[-m,m]{d_k}+[0,2]([-m,m]{d_k})^2)$ because:
$\frac{1}{1+t} \in 1-t+[0,2]t^2$ for any $t \ge -\frac{1}{2}$
Thus $d_{k+1} \in d_k - (d_k+[-m,m]{d_k}^2) (1+[-m,m]{d_k}+\frac{1}{3}[-m,m]d_k) \\ \quad \subseteq \frac{7}{3}[-m,m]{d_k}^2 + \frac{4}{3}[-m,m]^2{d_k}^3 \subseteq \frac{7}{3}[-m,m]{d_k}^2 + \frac{7}{18}[-m,m]{d_k}^2 \\ \quad \subset 3[-m,m]{d_k}^2 \subset [-1,1]d_k$
Thus the invariance is preserved
Also $3 m |d_{k+1}| < ( 3 m |d_k| )^2$
Therefore $3 m |d_k| < ( 3 m |d_0| )^{2^k} \le 2^{-2^k}$ for any natural $k$
Thus $x_k \to r$ quadratically as $k \to \infty$
Notes
In the case of finding $r = \sqrt[n]{a}$, the function $f : x \mapsto x^n - a$ has $\frac{f''}{2f'(r)}$ being $x \mapsto \frac{(n-1)x^{n-2}}{2r^{n-1}}$ which is bounded on $r[1,1+\frac{1}{4n})$ by $m = \frac{2n}{3r}$ because $\frac{n}{2r} (\frac{x}{r})^{n-2} \le \frac{n}{2r} (1+\frac{1}{4n})^n < \frac{n}{2r} e^{1/4} < m$. Thus $|x_0-r| < \frac{r}{4n} = \frac{1}{6m}$.
The procedure to find $x_0$ for efficient arbitrary precision arithmetic can be as follows:
Find the minimum integer $d$ such that $(2^d)^n \ge a$
Binary search on $[2^{d-1},2^d]$ to find $r$ until within an error of $\frac{2^{d-1}}{4n}$
Return the upper bound when the upper and lower bounds are within the error margin
The upper bound is between $r$ and $r+\frac{2^{d-1}}{4n} < r+\frac{r}{4n}$