Recursive Methods for Finding Roots of Functions

1. Recursive Methods for Finding Roots of Functions Bisection & Newton’s Method

2. Intermediate Value Theorem • Finding the root of a function will rely on this theorem. • It states that for any function that is continuous over the interval [a,b] then that function will take on every value between f(a) to f(b).

3. BISECTION • Bisection requires narrowing one root onto a certain interval [a,b] • Testing this interval in the function f should give values which are opposite in sign for f(a) and f(b). • Bisecting this interval and testing f((a+b)/2) will yield another smaller interval in which to continue bisecting until the desired accuracy is reached.

4. BisectionExample Given , find the root between 2 and 3 to an accuracy of 10^-3. • 1st Compare: f(2) = -1 and f(3) = 16. By the Intermediate Value Theorem we are insured that at least one root exists between 2 and 3. • 2nd Test f((2+3)/2) = f(2.5). All we need to know is it’s sign. It is positive. • 3rd, we narrow our search interval from [2,3] to [2,2.5] since f(2) is negative and f(2.5) is positive.

5. BisectionExample • Continuing this pattern to finish the problem: iteration 2: f(2.25)>0, new interval: [2,2.25] iteration 3: f(2.125)>0, new interval: [2,2.125] iteration 4: f(2.0625)<0, new interval: [2.0625,2.125] iteration 5: f(2.09375)<0, new interval: [2.09375,2.125] iteration 6: f(2.109375)>0, new interval: [2.09375,2.109375] iteration 7: f(2.1015625)>0, new int: [2.09375,2.1015625] iteration 8: f(2.09765625)>0, new int: [2.09375,2.09765625] iteration 9: f(2.095703125)>0, int: [2.09375,2.095703125] Root approximation is the midpoint of this interval: x=2.094390625

6. BisectionExample • We stop at this point because the desired accuracy has been reached. • The root is always as accurate as half of the interval arrived at, or in this case: (2.09503125-2.09375)/2 =.0009765625

7. BisectionAccuracy • Accuracy is always arrived at by taking half the current interval because the actual root could be no farther than that away from the midpoint. • The initial accuracy of the midpoint is: |b-a|/2 • After each iteration, this accuracy is halved. • After n iterations accuracy is: |b-a|/2^(n+1)

8. Psuedocode Input function, interval where root exists, and desired accuracy. {f, a, b, E} n = 1 While E > |b - a| / 2^n g = (b + a) / 2 //make a guess for root at midpoint if f(g) is the same sign as b then b = g else a = g endif n = n + 1 End loop Display guess as g to accuracy of |b - a| / 2^n

9. NEWTON’S METHOD • Requires you to make a relatively good guess for the root of f(x), x0. • Construct a tangent line to f(x) at this point: y - f(x0) = f´(x0)(x - x0) • Find the x-intercept of this line (y = 0): 0 - f(x0) = f´(x0)(x - x0) OR x = x0 - f(x0) / f´(x0) • Repeat guess for this new x.

10. Newton’s MethodExample Given , find the root between 2 and 3 to an accuracy of 10^-3. • 1st Guess: By view the graph make an initial guess of x0 = 2. • 2nd find new x: x1 = 2 - f(2) / f´(2) = 2.1 • 3rd repeat for x = 2.1

11. Newton’s MethodExample • Continuing this pattern takes much less time to narrow down than bisection: iteration 2: x2 = 2.1 - f(2.1) / f´(2.1) = 2.0945681211 iteration 3: x3 = x2 - f(x2) / f´(x2) = 2.0945514817 • You can already see that by the third iteration the accuracy of the root is to 0.0000166394 or less than 10^-3. • The number of iterations will depend on how good your guess is.

12. Psuedocode Input function, guess, and desired accuracy. {f, g, E} n = 0 While |gn - gn-1| > E n = n + 1 gn = gn-1 - f(gn-1) / f´(gn-1) End loop Display root as gn to accuracy of |gn - gn-1|

13. Quirks and Exceptions toNewton’s Method • if along the way f´(xn) = 0 then you will get a horizontal line with no x-intercept.

14. Quirks and Exceptions toNewton’s Method • You may get the wrong root depending on your initial guess. x0 x1

15. Quirks and Exceptions toNewton’s Method • You may get the wrong root. x0 x1