# Numerical Method -- 2

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Numerical Methods problems

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Request: Figure out the brackets and convergence and roots properly.

This may require careful thinking especially Q 1 case 1.

Part 1

Ignore roundoff error and assume infinite precision in all computations in this question.

We employ the bisection algorithm to find a root of the function f(x) graphed in Fig. 1 below.

1. DO NOT ATTEMPT to use a ruler to “measure the graph” to determine

the values of the roots, etc.

2. Solutions which attempt to “measure the graph” will not be considered valid.

We are given the following information about the function f(x):

1. The function f(x) is positive and never equals zero for x ≤ 0.

2. The function f(x) equals zero but does not cross zero at x = α, where 2 < α < 3.

3. The function f(x) is continuous everywhere except at x = 5.

4. The function f(x) is discontinuous and changes sign across x = 5 (= β). Also f(x) → ∞ as x → 5 from the left.

5. The function f(x) has a root in the interval 6 < x < 7. Denote this root by γ below.

6. The function f(x) equals zero but does not cross zero at x = 8 (= δ).

7. The function f(x) has a root in the interval 9 < x < 10. Denote this root by ε below.

8. The function f(x) is negative and never equals zero for x ≥ 10.

For each case below, state what will be the next step of the bisection algorithm.

It is essential that you answer this part of the question.

You will score zero for the whole question if you do not answer this part of the question.

1. xlow = 1.0, xhigh = 11.5.

2. xlow = 5.5, xhigh = 10.0.

3. xlow = 4.0, xhigh = 11.0.

4. xlow = 0.0, xhigh = 9.0.

5. xlow = 3.0, xhigh = 8.0.

6. xlow = 5.5, xhigh = 7.5.

7. xlow = 4.5, xhigh = 7.0.

8. xlow = −3000.0, xhigh = 9.0 (xlow not shown in Fig. 1). 9. xlow = 4.5, xhigh = 60.5 (xhigh not shown in Fig. 1).

For each case above which yields a valid initial bracket, to which root/discontinuity will the bisection algorithm converge?

“Convergence” is defined as follows in this question.

1. The tolerance for the bisection algorithm in this question is zero.

2. State the root/discontinuity the iterates will approach in the limit of an infinite number of iterations.

Part 2

• In this problem, the initial value of the low iterate is set to xlow = 4.0.

• We wish to find a value for xhigh such that the bisection algorithm converges to the root ε.

We know from Fig. 1 that ε lies in the interval 9 < ε < 10.

1. Note that if we select xhigh = ε, there will be no need for a bisection calculation.

However, since we do not know the value of ε in advance, this requires a lucky guess.

2. Explain why all choices for xhigh such that 5 ≤ xhigh < 9 can be rejected

instantly, regardless if we bracket a root/discontinuity or not.

3. Also explain why all choices for xhigh such that 4 < xhigh < 5 can be rejected instantly, regardless if we bracket a root/discontinuity or not.

4. Show that if we select xhigh ≥ 10, we will bracket a root or discontinuity, and there is a possibility that the bisection algorithm will converge to the root ε.

5. Next select xhigh such that 13 ≤ xhigh ≤ 14 and let xmid = (xlow + xhigh)/2.

(a) Calculate the range of values xmid takes, for xlow = 4 and 13 ≤ xhigh ≤ 14. (b) Which value of xlow or xhigh will be updated?

(c) How many roots of f(x) will be contained in the updated interval?

6. Explain why if we select xhigh such that 13 ≤ xhigh ≤ 14 (and xlow = 4), the bisection algorithm will converge to the root ε.

• “Convergence” is defined as follows in this question. 1. The tolerance on the bisection algorithm is zero.

2. State the root/discontinuity the iterates will approach in the limit of an infinite number of iterations.

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