27 julio 2010

Newton's Raphson Method





Newton's Raphson Method





 


In numerical analysis, Newton's method(also known as the Newton–Raphson method), named after Isaac Newton and Joseph Raphson, is perhaps the best known  method for finding successively better approximations to the zeroes (or roots) of a real-valued function. 
Newton's method can often  converge remarkably quickly, especially if the iteration begins "sufficiently near" the desired root. Just how near "sufficiently near" needs to be, and just how quickly "remarkably quickly" can be, depends on the problem. This is discussed in detail below. Unfortunately, when iteration begins far from the desired root, Newton's method can easily lead an unwary user astray with little warning. Thus, good implementations of the method embed it in a routine that also detects and perhaps overcomes possible convergence failures. Given a function ƒ(x) and its derivative ƒ '(x), we begin with a first guess x0.


 
Provided the function is reasonably well-behaved a better approximation
 x1 is 



The process is repeated until a sufficiently accurate value is reached:




 
An important and somewhat surprising application is Newton–Raphson division, which can be used to quickly find the reciprocal of a number using only multiplication and subtraction.
The algorithm is first in the class of Householder's methods, succeeded by Halley's method.


EXAMPLE

 Square root of a number



Consider the problem of finding the square root of a number. There are many methods of computing square roots, and Newton's method is one.


For example, if one wishes to find the square root of 612, this is equivalent to finding the solution to

\,x^2 = 612
The function to use in Newton's method is then,

\,f(x) = x^2 - 612
with derivative,

 f'(x) = 2x. \,
With an initial guess of 10, the sequence given by Newton's method is

\begin{matrix}
  x_1 & = & x_0 - \dfrac{f(x_0)}{f'(x_0)} & = & 10 - 
\dfrac{10^2 - 612}{2 \cdot 10} & = & 35.6 \quad\quad\quad{} \\
  x_2 & = & x_1 - \dfrac{f(x_1)}{f'(x_1)} & = & 35.6 - 
\dfrac{35.6^2 - 612}{2 \cdot 35.6} & = & \underline{2}6.3955056 
\\
  x_3 & = & \vdots & = & \vdots & = & 
\underline{24.7}906355 \\
  x_4 & = & \vdots & = & \vdots & = & 
\underline{24.7386}883 \\
  x_5 & = & \vdots & = & \vdots & = & 
\underline{24.7386338} 
\end{matrix}
Where the correct digits are underlined. With only a few iterations one can obtain a solution accurate to many decimal places.
  

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