Editing Newton's method (section)

===Slow convergence for roots of multiplicity greater than 1===
If the root being sought has [[Multiplicity (mathematics)#Multiplicity of a root of a polynomial|multiplicity]] greater than one, the convergence rate is merely linear (errors reduced by a constant factor at each step) unless special steps are taken. When there are two or more roots that are close together then it may take many iterations before the iterates get close enough to one of them for the quadratic convergence to be apparent. However, if the multiplicity {{mvar|m}} of the root is known, the following modified algorithm preserves the quadratic convergence rate:<ref>{{cite web|title=Accelerated and Modified Newton Methods|url=http://mathfaculty.fullerton.edu/mathews/n2003/newtonacceleratemod.html|access-date=4 March 2016|archive-url=https://web.archive.org/web/20190524083302/http://mathfaculty.fullerton.edu/mathews/n2003/NewtonAccelerateMod.html|archive-date=24 May 2019|url-status=dead}}</ref>

<math display="block">x_{n+1} = x_n - m\frac{f(x_n)}{f'(x_n)}. </math>

This is equivalent to using [[successive over-relaxation#Other applications of the method|successive over-relaxation]]. On the other hand, if the multiplicity {{mvar|m}} of the root is not known, it is possible to estimate {{mvar|m}} after carrying out one or two iterations, and then use that value to increase the rate of convergence.

If the multiplicity {{mvar|m}} of the root is finite then {{math|1={{var|g}}({{var|x}}) = {{sfrac|{{var|f}}({{var|x}})|{{var|{{prime|f}}}}({{var|x}})}}}} will have a root at the same location with multiplicity 1.  Applying Newton's method to find the root of {{math|{{var|g}}({{var|x}})}} recovers quadratic convergence in many cases although it generally involves the second derivative of {{math|{{var|f}}({{var|x}})}}.  In a particularly simple case, if {{math|1={{var|f}}({{var|x}}) = {{var|x}}{{sup|{{var|m}}}}}} then {{math|1={{var|g}}({{var|x}}) = {{sfrac|{{var|x}}|{{var|m}}}}}} and Newton's method finds the root in a single iteration with

<math display="block">x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)} = x_n - \frac{\;\frac{x_n}{m}\;}{\frac{1}{m}} = 0\,.</math>