Newton’s Iterative Method Newton’s iterative method, widely known as the Newton-Raphson method, is an algorithmic tool used in numerical analysis to approximate the real roots (or zeroes) of a differentiable function. The technique starts with an initial numerical guess and applies a calculus-based formula repeatedly to generate increasingly accurate approximations. 1. Mathematical Derivation
The primary objective is to find a solution to the equation: f(x)=0f of x equals 0
is a continuous and differentiable function, we start with an initial guess, denoted as . To find a better approximation , we evaluate the first-order Taylor series expansion of
f(x)≈f(x0)+f′(x0)(x−x0)f of x is approximately equal to f of open paren x sub 0 close paren plus f prime of open paren x sub 0 close paren open paren x minus x sub 0 close paren
Setting this linear approximation equal to zero to find its intercept yields:
f(x0)+f′(x0)(x1−x0)=0f of open paren x sub 0 close paren plus f prime of open paren x sub 0 close paren open paren x sub 1 minus x sub 0 close paren equals 0 Solving for gives the foundational step of the algorithm:
x1=x0−f(x0)f′(x0)x sub 1 equals x sub 0 minus the fraction with numerator f of open paren x sub 0 close paren and denominator f prime of open paren x sub 0 close paren end-fraction Generalizing this for any step results in the standard iterative formula:
xn+1=xn−f(xn)f′(xn)x sub n plus 1 end-sub equals x sub n minus the fraction with numerator f of open paren x sub n close paren and denominator f prime of open paren x sub n close paren end-fraction 2. Geometric Interpretation
Geometrically, Newton’s method works by replacing a complex curve with its local linear approximation. Evaluate Point: The algorithm finds the point on the graph of the function.
Construct Tangent: It constructs a tangent line to the curve at that exact point. The slope of this line is given by the derivative Find Intercept: The next approximation, xn+1x sub n plus 1 end-sub
, is defined as the exact point where this tangent line intersects the x-axis.
This sequence repeats, with each successive tangent line bringing the x-intercept closer to the actual root of the function. 3. Step-by-Step Example Consider approximating the square root of
, which is equivalent to finding the positive root of the polynomial function: f(x)=x2−2=0f of x equals x squared minus 2 equals 0 Step 1: Compute the Derivative Differentiating with respect to f′(x)=2xf prime of x equals 2 x Step 2: Establish the Iterative Formula Substituting into the Newton-Raphson formula gives:
xn+1=xn−xn2−22xnx sub n plus 1 end-sub equals x sub n minus the fraction with numerator x sub n squared minus 2 and denominator 2 x sub n end-fraction Simplifying the algebraic fraction:
xn+1=12(xn+2xn)x sub n plus 1 end-sub equals one-half open paren x sub n plus the fraction with numerator 2 and denominator x sub n end-fraction close paren Step 3: Run the Iterations Using an initial guess of Iteration ( Current Approximation ( 1.4166666671.416666667 0.0069444440.006944444 1.4142156861.414215686 0.0000060070.000006007 1.4142135621.414213562 0.0000000000.000000000 4. Convergence and Limitations
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