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Newton-Raphson method is used to find the roots/zero of real valued function.

I am not able to understand the correctness of the algorithm.

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$

my question is why the equation guarantees that it will eventually find the root of function after multiple iterations.

EdG
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Prashant Bhanarkar
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  • see here http://www.youtube.com/watch?v=oE98W4A7Zio i hope this will help you – Dr. Sonnhard Graubner May 31 '17 at 17:52
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    The equation does not guarantee this. It only does it for values of $x_0$ close enough to a root, assuming $f$ is reasonably nice. – Andrés E. Caicedo May 31 '17 at 17:53
  • Some conditions are required for the method to converges, nonetheless, the idea comes from: $$f(x)=f(x_0)+f'(x)(x-x_0)+o(x-x_0).$$ Instead of seeking the zeroes of a non-linear function, one uses the linear estimate of this function, namely its tangent linear map. – C. Falcon May 31 '17 at 17:54
  • It doesn't guarantee, but it usually does work. The premise is that all continuous functions are approximately linear at some level. So we use a linear approximation, find the zero for that line and repeat. – Doug M May 31 '17 at 17:54
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    "[...] guarantees [...]" It doesn't. $f$ might not even have a root. – Chappers May 31 '17 at 17:57
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    The formula can be deduced fast remembering that it finds the root in one iteration if the function is a straight line. $f'(x_n)=\dfrac{f(x_{n+1})-f(x_n)}{x_{n+1}-x_{n}}$ with $f(x_{n+1})=0$ – Jaume Oliver Lafont May 31 '17 at 18:13
  • In general, the method does not find roots, especially not in a finite number of steps. In the case that the method works, it finds successively better approximations of a root. – Lutz Lehmann Jun 01 '17 at 07:57

1 Answers1

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Here's A DIY approach. Grab a pencil and paper

(0) Draw a curve intersecting the $x$-axis at some point.

(1) Mark a point $x_o$ on the $x$-axis at some initial guess for $x$. Set $k=0$

(2) Draw a vertical line from $x_k$ on the $x$-axis till it meets the curve.

(3) From this point on the curve draw a tangent to the curve, till it intersects the $x$-axis.

(4) Set $k=k+1$ and Call this new point on the $x$-axis, $x_k$ (This is your updated value) .

Repeat Steps (2) -- (4) till convergence

It is ONLY guaranteed to work if $f(x)$ has a continuous first derivative and no saddle point (zero slope with $f(x)\ne 0$) between your initial guess and the root closest to the initial guess, if it exists.

Sharat V Chandrasekhar
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