NIELIT 2022 April Scientist B | Section B | Question: 88

Question

The secant method formula for finding the square root of a real number $\text{R}$ from the equation $x^{2} – \text{R} = 0$ is:

• A. $x_{i+1} = \frac{x_{i}.x_{i-1}}{x_{i} + x_{i+1}}$
• B. $x_{i+1} = \frac{x_{i}.x_{i-1} + \text{R}}{x_{i} + x_{i-1}}$
• C. $x_{i+1} = \frac{1}{2}\left(x_{i} + \frac{\text{R}}{x_{i}}\right)$
• D. $x_{i+1} = \frac{2x_{i}^{2} + x_{i}.x_{i-1} – \text{R}}{x_{i} + x_{i-1}}$ 

Answer 1

Let us understand first Theory to the problem and it is basic application of derivatives.

our problem is to find the square root of a real number R.

So,

$\large x=\sqrt{R}$

$\large x^{2}={R}$

$\large x^{2}-{R}=0$

say , $\large f(x)=x^{2}-{R}$

so we need to find the solution when f(x)=0.

Idea:-

let’s see the idea through the graph.

Lets assume a point $\large x_{0}$ is the required solution of our equation $\large x^{2}-{R}=0$.

$\large x_{N}$ is the desired point which is square root of R.

(say for practical purpose if you want to find the square root of 5 you will choose $\large x_{0}$=2 as we know it is the closest assumption to square root of 5).

now the corresponding y co-ordinate is $\large x_{0}^{2}-R$.

Now we will draw a tangent on $\large \left ( x_{0},x_{0}^{2}-R \right )$  as we show in the diagram it will intersect the x axis at point $x_{1}$ .

Now if you observe carefully in the diagram the point $x_{1}$ is closer to the point $x_{N}$ than $x_{0}$.

Now if we again do the same analysis again by drawing a tangent from $\large \left ( x_{1},x_{1}^{2}-R \right )$ we will get $x_{2}$ which will be more closer to $x_{N}$ than $x_{1}$.

This is the crux of this method  if we do this procedure 4-5 times we will get an accurate answer .

But this geometric procedure is not feasible to write a program for this problem so we have to derive some mathematical formula .

now we know ,

the equation of the tangent is,

$\large (y-y_{0})=m(x-x_{0})$

m=slope of the tangent=derivative of the function

$\large m=f'(x)=2x$

we will put y=0 in the equation as the tangent will intersect with x axis and that point y=0.

$\large -y_{0}=2x_{0}(x-x_{0})$           [as we will draw tangent at ($x_{0},y_{0}$) that point slope=2$x_{0}$]

$\large x=x_{0}-\frac{y_{0}}{2x_{0}}$

$\large x=x_{0}-\frac{x_{0}^{2}-R}{2x_{0}}$        [as $\large y_{0}=x_{0}^{2}-R$]

$\large x=\frac{x_{0}}{2}+\frac{R}{2x_{0}}$

so we got x1 by this equation.

The formula looks like ,

$\large x_{i+1}=\frac{x_{i}}{2}+\frac{R}{2x_{i}}$

correct answer is C.

useful links:-

1. https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/fb8c24f09aba8413984f8ce5961586bd_lec13.pdf

2. https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/resources/lecture-13-newtons-method/