Newton’s method
- Also called:
- Newton-Raphson method
Newton’s method, a mathematical technique for solving equations involving a polynomial expression being equal to zero—that is, f(x) = 0. The method uses successive approximations to find a value of x that best gives a value of zero in the polynomial expression. The method was devised by Isaac Newton in 1669 on the basis of a method of French mathematician Francois Viète (who may in turn have learned of it from the work of Persian astronomer al-Kāshī). In 1690 English mathematician Joseph Raphson greatly simplified Newton’s method to its current form.
The method
The method starts with an initial guess, usually labeled x0, such that f(x0) is close to zero. The tangent line to the graph of y = f(x) is then found and traced back to the x-axis to find a new and better guess, x1, to f(x) = 0. This process can be represented asx1 = x0f(x0)/f′(x0),where f′(x0) is the derivative (or tangent slope) of the function f(x) at the initial guess x0.
This process is then repeated to find a new and better guess by finding the equation of the tangent line at the value x = x1 and tracing this back to the x-axis to find a better guess, x2. This step can be represented asx2 = x1f(x1)/f′(x1),where f′(x1) is the derivative (or tangent slope) of the function f(x) at the initial guess x1.
This process is repeated until a value for x is obtained that is close enough for the application of the process to the given situation. A general recursive formula for the process can be written asxn + 1 = xnf(xn)/f′(xn).
Example
One can use Newton’s method to find the square root of 5—that is, x2 − 5 = 0. Thus, f(x) = x2 − 5 and f′(x) = 2x. If the initial guess x0 = 2 is used to start the solution, f(x0) = 22 − 5 = −1, and f′(x0) = 2(2) = 4, thenx1 = 2 − −1/4 = 2.25.
To check this new solution, one begins with f(x1) = f(2.25) = 2.252 − 5 = 0.0625, which means that x = 2.25 is much closer to the square root of 5 than the initial guess of 2 is. If one applies the recursive formula again, with f(x1) = 2.252 − 5 = 0.0625 and f′(x 0) = 2(2.25) = 4.5, then x2 = 2.25 − 0.0625/4.5 = 2.2361,which is the square root of 5 to four decimal places. The process can be repeated for even greater accuracy.