روش نصف کردن. Bisection Method Basis of Bisection Method

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1 10/18/01 روش نصف کردن Bisection Method 1 Basis o Bisection Method Theore An equation ()=0, where () is a real continuous unction, has at least one root between l and u i ( l ) ( u ) < 0. () u Figure 1 At least one root eists between the two points i the unction is real, continuous, and changes sign. 1

2 10/18/01 Basis o Bisection Method () u Figure I unction does not change sign between two points, roots o the equation 0 ay still eist between the two points. Basis o Bisection Method () () u u Figure I the unction does not change sign between two points, there ay not be any roots or the equation 0 between the two points. 4

3 10/18/01 Basis o Bisection Method () u Figure 4 I the unction changes sign between two points, ore than one root or the equation 0 ay eist between the two points. 5 Algorith or Bisection Method 6

4 10/18/01 Step 1 Choose and u as two guesses or the root such that ( ) ( u ) < 0, or in other words, () changes sign between and u. This was deonstrated in Figure 1. () u Figure 1 7 Step Estiate the root, o the equation () = 0 as the id point between and u as () = u u Figure 5 Estiate o 8 4

5 10/18/01 Step Now check the ollowing 0 a) I l, then the root lies between and ; then = ; u =. 0 b) I l, then the root lies between and u ; then = ; u = u. 0 c) I l ; then the root is. Stop the algorith i this is true. 9 Step 4 Find the new estiate o the root = u Find the absolute relative approiate error where a old new new old new 100 previousestiateo root current estiateo root

6 10/18/01 Step 5 Copare the absolute relative approiate error the pre-speciied error tolerance. Is a s? Yes s a Go to Step using new upper and lower guesses. with No Stop the algorith Note one should also check whether the nuber o iterations is ore than the aiu nuber o iterations allowed. I so, one needs to terinate the algorith and notiy the user about it. 11 Eaple 1 You are working or DOWN THE TOILET COMPANY that akes loats or ABC coodes. The loating ball has a speciic gravity o 0.6 and has a radius o 5.5 c. You are asked to ind the depth to which the ball is suberged when loating in water. Figure 6 Diagra o the loating ball 1 6

7 10/18/01 Eaple 1 Cont. The equation that gives the depth to which the ball is suberged under water is given by a) Use the bisection ethod o inding roots o equations to ind the depth to which the ball is suberged under water. Conduct three iterations to estiate the root o the above equation. b) Find the absolute relative approiate error at the end o each iteration, and the nuber o signiicant digits at least correct at the end o each iteration. 4 1 Eaple 1 Cont. Fro the physics o the proble, the ball would be suberged between = 0 and = R, where R = radius o the ball, that is 0 R Figure 6 Diagra o the loating ball

8 10/18/01 Eaple 1 Cont. To aid in the understanding o how this ethod works to ind the root o an equation, the graph o () is shown to the right, where Solution Figure 7 Graph o the unction () 15 Eaple 1 Cont. Let us assue u Check i the unction changes sign between and u. 4 4 l u Hence l u So there is at least on root between and u, that is between 0 and

9 10/18/01 Eaple 1 Cont. Figure 8 Graph deonstrating sign change between initial liits 17 Iteration 1 The estiate o the root is Eaple 1 Cont u l Hence the root is bracketed between and u, that is, between and So, the lower and upper liits o the new bracket are l 0.055, 0.11 u At this point, the absolute relative approiate error calculated as we do not have a previous approiation. a cannot be

10 10/18/01 Eaple 1 Cont. Figure 9 Estiate o the root or Iteration 1 19 Eaple 1 Cont u (0.085) l Iteration The estiate o the root is Hence the root is bracketed between and, that is, between and So, the lower and upper liits o the new bracket are l 0.055, u

11 10/18/01 Eaple 1 Cont. Figure 10 Estiate o the root or Iteration 1 Eaple 1 Cont. The absolute relative approiate error a at the end o Iteration is a new new old % None o the signiicant digits are at least correct in the estiate root o = because the absolute relative approiate error is greater than 5%. 11

12 10/18/01 Eaple 1 Cont u l Iteration The estiate o the root is Hence the root is bracketed between and, that is, between and So, the lower and upper liits o the new bracket are l 0.055, u Eaple 1 Cont. Figure 11 Estiate o the root or Iteration 4 1

13 10/18/01 Eaple 1 Cont. The absolute relative approiate error a at the end o Iteration is a new new old % Still none o the signiicant digits are at least correct in the estiated root o the equation as the absolute relative approiate error is greater than 5%. Seven ore iterations were conducted and these iterations are shown in Table 1. 5 Table 1 Cont. Table 1 Root o ()=0 as unction o nuber o iterations or bisection ethod. Iteration u a % ( )

14 10/18/01 Table 1 Cont. Hence the nuber o signiicant digits at least correct is given by the largest value or or which log 0.44 a log So The nuber o signiicant digits at least correct in the estiated root o at the end o the 10 th iteration is. 7 Always convergent Advantages The root bracket gets halved with each iteration - guaranteed

15 10/18/01 Slow convergence Drawbacks I one o the initial guesses is close to the root, the convergence is slower 9 Drawbacks (continued) I a unction () is such that it just touches the -ais it will be unable to ind the lower and upper guesses. ()

16 10/18/01 Drawbacks (continued) Function changes sign but root does not eist () 1 1 THE END 16