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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 \( x_0\), find the value of \( f(x_0) \), take the tangent to the curve, which results in the approximate root \(x_1\). We continue this process, find the value of \( f(x_1) \), take the tangent to the curve, which results in the approximate root \(x_2 \). By continuing the process, we should reach the actual root \( x^* \), 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 \( x_0 \). Then, we consider the point \( (x_0, f(x_0) \), which has a slope of \( f'(x_0) \), hence the equation of the tangent is

\[ \frac{ y - f(x_0) } { x - x_0 } = f' (x_0) \]

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

\[ x_1 = x_0 - \frac{ f(x_0)} { f'(x_0)} . \]

We repeat this process with

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

until a sufficiently accurate value is reached.

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

Note by Chung Gene Keun
2 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 · 2 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 · 2 years, 10 months ago

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@Eddie The Head 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 · 2 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 · 2 years, 8 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 · 2 years, 1 month ago

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