3.8: Newton’s Method Greg Kelly, Hanford High School, Richland, Washington
Objective Approximate a zero of a function using Newton's Method.
Interval Midpoint(c) f(c) Recall the Bisection Method for approximating roots. Need interval where f(endpoints) have different signs. Interval [ -, +] Midpoint(c) f(c) [1, 2] 1.5 + [1, 1.5] 1.25 - [1.25, 1.5] 1.375 [1.375, 1.5] 1.4375 Converges, but it may take a while.
Another method for finding zeros (besides the Bisection Method) is called Newton’s Method (also known as Newton-Raphson Method). Requirements for Newton’s Method: Needs an initial “good” estimate Uses the derivative of the function in the calculation, so the function needs to be differentiable in the open interval (a,b) around the zero
Newton's Method is based on the assumption that the graph of f and the tangent line at (x,f(x)) both cross the x-axis at about the same point. You use that x-intercept to make another estimate (usually better) for the zero.
Tangent line: set y=0 and solve for x Newton's Method: Tangent line: set y=0 and solve for x intercept new guess Next estimates:
Using Newton's Method: Let f(c)=0 where f is differentiable on an open interval containing c. Make an initial estimate x1 (that's "close" to c). Determine a new approximation: If is within the desired accuracy, stop. Otherwise repeat step 2.
Calculate 3 iterations of Newton's Method to approximate a zero of Use x1=1 as the initial guess. Find x4.
TI 84 Procedure: Set y1 = function and y2 = derivative Type in the initial approximation and hit ENTER. Type in ANS – y1(ANS) / y2(ANS) hit ENTER Hit ENTER for next approximation. Repeat step 4 until the desired accuracy is reached.
Use Newton's Method to approximate the zeros of Continue until successive approximations differ by less than 0.0001. Test some #s: Or look at graph on calculator (x≈-1.2) x1=-1.2 x1=-1. x1=0 x2=-1.23511 x2=-1.333333 x2=1 x3=-1.233754 x3=-1.243386 x3=0.5715286 x4=-1.23375 x4=-1.233855 x5=-1.23375 x6=-1.23375 x18=-1.23375
Use Newton's Method to find zeros of Let x1=0.1.
There are some limitations to Newton’s method: Looking for this root. Bad guess. Wrong root found Failure to converge
Condition for Convergence:
Example: On the interval (1,3), it will be <1. So convergence of Newton's method is guaranteed.
Example: For any value of x, it equals 2, so Newton's Method will fail to converge.
Homework 3.8 (p. 226) #1-17 odd 27, 29, 33 (#5-13 only find one root)