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I don't quite understand what is the difference between approximating the first positive root of a function and third positive root of a function by using the Newton-Raphson formula. Can the following formula apply to finding the third positive root ?

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

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    $\begingroup$ Newton's method finds a root; which one depends on where you start. $\endgroup$ – Paul Sep 15 '16 at 9:26
  • $\begingroup$ @Paul Thanks for help! do you mean it's depends the initial condition I picked up ? $\endgroup$ – phantom Sep 15 '16 at 9:28
  • $\begingroup$ Yes - see answer below. $\endgroup$ – Paul Sep 15 '16 at 9:29
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Yes, Newton's method works by assuming that if you were to plot a graph of a function and choose a point near to a root of that function then a lime drawn tangential to the curve to the x-axis will be closer to the root and we repeat this to get better and estimates.

This does not have to be the first root. If your starting point is near enough to another root you should get that root.

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See the pretty pictures here. Which root the algorithm converges too depends on the starting point. So yes, you can find the 3th root, but you'll have to make sure you start at the right point.

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The Newton-Raphson iteration formula can converge towards any root of the polynomial depending on where you start it.

If $x_1$ is close to one of the roots, then the iteration will generally converge towards that root.

If you want to find a particular root, you'll have to use different techniques to get an initial estimate of its location so you can start off the iteration at a good point.

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