Newton Raphson method of root aproximation

The Newton Raphson method finds successively better approximations to roots of the real valued function \( f(x) = 0 \).

Refer to the above image. We start off with the approximate root x0 x_0, find the value of f(x0) f(x_0) , take the tangent to the curve, which results in the approximate root x1x_1. We continue this process, find the value of f(x1) f(x_1) , take the tangent to the curve, which results in the approximate root x2x_2 . By continuing the process, we should reach the actual root x x^* , to a great degree of accuracy.

Let's write this down in equation form. We start off with a first approximation, which is denoted by x0 x_0 . Then, we consider the point (x0,f(x0) (x_0, f(x_0) , which has a slope of f(x0) f'(x_0) , hence the equation of the tangent is

yf(x0)xx0=f(x0) \frac{ y - f(x_0) } { x - x_0 } = f' (x_0)

and the intersection with the x-axis occurs when y=0 y=0 , or that

x1=x0f(x0)f(x0). x_1 = x_0 - \frac{ f(x_0)} { f'(x_0)} .

We repeat this process with

xn+1=xnf(xn)f(xn), x_{n+1} = x_n - \frac{ f(x_n) } { f'(x_n) },

until a sufficiently accurate value is reached.

Find the value of 2 \sqrt{2} as the root of the equation f(x)=x22 f(x) = x^2 -2 , using the starting value of x0=2 x_0 = 2 . Note that all values of xnx_n will be rational numbers, hence we get a sequence of rationals which approximate the value of the irrational number 2 \sqrt{2} .

Note by Chung Gene Keun
6 years, 10 months ago

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Ahh!? Finally, there is someone here make a note about this topic. Now, using the Newton-Raphson method, try to answer these problems:

  1. Simultaneous Linear Equations

  2. Simultaneous Non-Linear Equations

  3. Computational Physics (Mechanical System)

Good luck!! (^_^)

Tunk-Fey Ariawan - 6 years, 10 months ago

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Ugh!!I wanted to make a note on this topic....but you beat me to it!!Nicely presented btw!!

Eddie The Head - 6 years, 10 months ago

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Thanks! I wanted to create problems on the Newton Raphson method, and highlight the cases when it failed, as those were interesting to me.

I found that explaining this method in the problem made it too long, so I separated it out into a note, and will link it from there. Problems should be posted soon :)

Chung Gene Keun - 6 years, 10 months ago

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Here, f(x) = x^2 - 2 and f’(x) = 2x

Now, f(0)=-2




hence, the root lies between 1 and 2.

Let, x0 = 1

x1 = x0 – { f(x0) / f’(x0)}

 = 1 – {(-1) / 2}

 = 1.5


x2 = 1.4166

x3 = 1.4142

x4 = 1.4142

hence, √2 = 1.4142

Debika Sadhukhan - 6 years, 8 months ago

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Santhosh HD - 6 years, 4 months ago

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Isnt it better to use it to approximate roots of higher degree equations than finding square roots?

Deepak Pant - 6 years, 1 month ago

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