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. Bisection Example Given, find the root between 2 and 3 to an accuracy of 10^-3. §3rd, we narrow our search interval from [2,3] to [2,2.5] since f(2) is negative and f(2.5) is positive. §2nd Test f((2+3)/2) = f(2.5). All we need to know is it’s sign. It is positive. §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.

5. §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 Bisection Example

6. §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 Bisection Example

7. Bisection Accuracy §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), x 0. §Construct a tangent line to f(x) at this point: y - f(x 0 ) = f´(x 0 )(x - x 0 ) §Find the x-intercept of this line (y = 0): 0 - f(x 0 ) = f´(x 0 )(x - x 0 ) OR x = x 0 - f(x 0 ) / f´(x 0 ) §Repeat guess for this new x.

10. Newton’s Method Example 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 x 0 = 2. §2nd find new x: x 1 = 2 - f(2) / f´(2) = 2.1 §3rd repeat for x = 2.1

11. §Continuing this pattern takes much less time to narrow down than bisection: iteration 2: x 2 = 2.1 - f(2.1) / f´(2.1) = 2.0945681211 iteration 3: x 3 = x 2 - f(x 2 ) / f´(x 2 ) = 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. Newton’s Method Example

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

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

14. Quirks and Exceptions to Newton’s Method §You may get the wrong root depending on your initial guess. x0x0 x1x1

15. Quirks and Exceptions to Newton’s Method §You may get the wrong root. x0x0 x1x1