Today s class. Numerical differentiation Roots of equation Bracketing methods. Numerical Methods, Fall 2011 Lecture 4. Prof. Jinbo Bi CSE, UConn

Transcription

1 Today s class Numerical differentiation Roots of equation Bracketing methods 1

2 Numerical Differentiation Finite divided difference First forward difference First backward difference Lecture 3 2

3 Numerical Differentiation Centered difference approximation Subtract the two equations Lecture 3 3

4 Numerical Differentiation First forward difference Lecture 3 4

5 Numerical Differentiation First backward difference Lecture 3 5

6 Numerical Differentiation Centered difference Lecture 3 6

7 Error Propagation What is the effect of error in one calculation propagating to subsequent calculations? Example: Multiplying sin x with cos x Single variable functions Lecture 3 7

8 Error Propagation Use Taylor series Lecture 3 8

9 Error Propagation Lecture 3 9

10 Error Propagation Multivariable functions Lecture 3 10

11 Numerical stability Condition of a problem is a measure of its sensitivity to changes in input values The condition number is defined as the ration of the relative function error to the relative value error Lecture 3 11

12 Numerical stability Condition number < 1 indicates a wellconditioned function I.e. changes in the input are attenuated Condition number > 1indicates a illconditioned function I.e. changes in the input are amplified Lecture 3 12

13 Roots of equation Given a function f(x), the roots are those values of x that satisfy the relation f(x) = 0 Example From the quadratic formula, the roots are: 13

14 Roots of Equations The need to solve for roots show up in many engineering problems Also, can be used to find solutions to implicit variables 14

15 Example Find a value of R such that current is 5A at t = 1s 15

16 Example It is not possible to isolate R to the left side and thus solve for R R is know as an implicit variable Rewrite the function as a function of R set to 0 16

17 Roots of equations Still need a method to solve for this root Other examples of difficult to solve roots 17

18 Roots of equations Non-computer methods Graphical methods 18

19 Graphical methods Not exact Can give you a rough estimate of the root, Can give you insights on the number of roots and shape of the curve Can use the rough estimate in more precise numerical methods 19

20 Graphical methods Use to get an initial estimate of the root and also to find out how many roots there are 20

21 Graphical methods 21

22 Graphical methods 22

23 Graphical methods 23

24 Roots of equation Non-computer method Exhaustive search method To find the root in the interval [a,b], start at x=a and check if f(a) = 0, then try f(a+δ), f(a+2δ), and so on, until we get f(x) sufficiently close to 0 If the step value Δ is sufficiently small we can obtain an accurate result but this could take an extremely long time. For example, if the interval is [0,10] and the step size is Δ = 0.001, it will take on average 10,000 guesses In addition to the inefficiency of this approach, if f(x) is a steep function, this approach may not produce an accurate results 24

25 Exhaustive search Example Find the root of the function Actual root is at x= With an interval of [0.9, 1.1] and a step size of Δ = The exhaustive search method will test f(1.000) = and f(1.001) = 0.086, neither of which are that close to f(x) = 0 25

26 Roots of equations More systematic methods are required Bracketing methods Open methods 26

27 Incremental search methods Locate an interval where sign changes Divide interval into smaller subintervals which are then searched for sign changes Keep repeating until root is found with sufficient confidence 27

28 Bisection method Also called: Binary chopping Interval halving An incremental search method where the interval is cut in half 28

29 Bisection method Step 1: Choose lower x l and upper x u such that the function changes sign over that range i.e. f(x l ) and f(x u ) are different signs or f(x l ) f(x u ) < 0 Step 2: Estimate root to be x r =(x l +x u )/2 29

30 Bisection method Step 3: Determine in which subinterval the root lies If f(x r ) 0 is within acceptable tolerance, stop and root equals x r If f(x l ) f(x r ) < 0, then root is in lower subinterval. Set x u = x r, and return to step 2 If f(x l ) f(x r ) > 0, then root is in upper subinterval. Set x l = x r, and return to step 2 30

31 Bisection method Termination criteria Use approximate relative error calculation to determine when to stop In general, ε a is larger than ε t 31

32 Bisection method Example: Use range of [202:204] Root is in upper subinterval 32

33 Bisection method Use range of [203:204] Root is in lower subinterval 33

34 Bisection method Use range of [203:203.5] Root is in upper subinterval 34

35 Error estimates The approximate error is upper bound estimate of the true error When the root is near one of the ends of the interval, the approximate error is fairly close to the actual true error Error is fairly well-contained 35

36 Error estimates You always know that the true root is within Δx/2 of your estimate 36

37 Bisection method 37

38 Bisection method You can calculate an error estimate based just on the initial guesses You can also make estimates on the error on future iterations Superscripts indicates the iteration number 38

39 Bisection method Each subsequent iteration cuts the approximate error in half This, allows to determine a priori exactly how many iterations are needed to arrive at the desired error 39

40 False Position Method The false position method works in a similar fashion to the bisection method Start with an initial interval [a,b] where f(a) and f(b) have opposite signs, which is the same as the bisection Instead of choosing the initial guess x r as the midpoint of the interval, we join the point {a,f(a)} and {b,f(b)} with a straight line and choose x r as the point where that straight line crosses the x-axis. 40

41 False Position Method 41

42 False Position Method Algorithm is the same as bisection method with the same three steps 42

43 False Position Method Step 1: Choose lower x l and upper x u such that the function changes sign over that range - i.e. f(x l ) and f(x u ) are different signs - or f(x l ) f(x u ) < 0 Step 2: Estimate new root to be 43

44 False Position Method Step 3: Determine in which subinterval the root lies If f(x r ) 0 is within acceptable tolerance, stop and root equals x r If f(x l ) f(x r ) < 0, then root is in lower subinterval. Set x u = x r, and return to step 2 If f(x l ) f(x r ) > 0, then root is in upper subinterval. Set x l = x r, and return to step 2 44

45 Next class Roots of equations Open methods Read chapters 5 and 6 HW2, due 9/15 Chapra & Canale 6 th edition 3.5, 3.7, 3.13, 4.5, 4.6, 4.12 (b) and (d), and