Emth271 12 Exam

8/12/2019 Emth271 12 Exam

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Surname/Family Name: _______________________________

First Name/Given Name: _______________________________

Student ID: __________________________________________

Seat Number: _____________________________________

University of Canterbury

END OF YEAR EXAMINATIONS 2012

Time Allowed: TWO hours

Number of Pages: 34

Prescription Number(s): EMTH271-12S2 / MATH270-12S2

Paper Title: Mathematical Modelling andComputation 2

MARKFOR OFFICE USE ONLY

Q1

Q2

Q3

Q4

Q5

TOTAL

Read these instructions carefully:

• Attempt ALL FIVE questions.

• Write your answers in the spaces provided.

• You may use the left-hand pages for rough working.

• All questions are worth equal marks.

• You are permitted one A4 sheet of notes in the exam.

• Calculators are permitted.

• Use black or blue ink only (not pencil).

• Explain everything and show all working.

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Use this page for rough working

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1. (a) Find the roots of f (x) = x2 − 5x + 6 and verify that f (x) = 0 is equivalent tox = g(x) for each of the following choices:

i. g(x) = (5x − 6)1/2.

ii. g(x) = (x2 + 6)/5.(b) Show that the iterative methods in (i) and (ii) converge to only one root each

for suitable starting values. State all the theorems and results you use.

TURNOVER

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(c) Let · be a norm, e.g. · 2 or · ∞. Let A be a non-singular n × n matrix.Define the condition number, κ(A), of A.

(d) Let

An = 1 22 4 + 1

n2i. Find A−1

n .

ii. Using infinity norm, find κ(An).

iii. What happens to κ(An) as n → ∞?

iv. Given the information above, if you were to solve the system Anx = b

numerically for large n, would you trust your solution? Explain.

TURNOVER

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(e) Considerx -1 0 2 3f (x) -1 3 11 27

estimate f (1) as accurately as possible.

TURNOVER

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ii. Using the fact that

M 2J =

9/14 0

0 9/14

and that M

4J = M

2J M

2J , M

6J = M

2J M

2J M

2J , etc., show that the entries of

M 2nJ all tend to zero as n → ∞.

TURNOVER

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iii. Using (i) and (ii) give a bound for ||M kJ ||∞ for any k.

iv. Using the fact that errors for the iterative method satisfy

e(k+1) = M J e

(k),

show that Jacobi’s method converges as k → ∞ even though A is notdiagonally dominant.

TURNOVER

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(b) Consider the differential equation

d2y

dx2 +

dy

dx + 2xy = (1 − x2).

Now boundary conditions are specified as:

y(0) = 1, y(1) = 2.

Let xi = ih and yi = y(xi) for i = 0, . . . , N , where h = 1/N is the step size.

i. Use central differences on the differential equation to write down an equa-tion involving yi−1, yi and yi+1.

TURNOVER

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ii. Choose N=3. Then use the boundary conditions together with your lastequation to write down the two equations in the unknown yi’s, namely fori = 1 and i = 2.

iii. Now write these equations as a matrix system for this problem with N = 3.

TURNOVER

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3. A quadrature formula to approximate the definite integral

I (f ) =

b

a

f (x) dx

associated with the integrand f (x). This quadrature rule, denoted by Q, is definedby

Q(f ) =n

i=0

wif (xi)

where here n = 2. The integrand f (x) is to be evaluated at the uniform nodesa = x0 < x1 < x2 . . . < xn = b with uniform step size h such that xi+1 = xi + h,i=0(1)n, with h=(b-a)/n.

It is desired to find a quadrature formula with precision at least 2.

(a) By the method of undetermined coefficients show how this leads to a linearsystem for the weights wi. Write this system down in matrix form

(b) Explain any potential difficulties with this method of finding the weights. Listat least two.

TURNOVER

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(c) The composite Simpson rule was used to integrate

I (f ) = 1

0

f (x) dx,

when the integrand was1

1 + x4

and the following results shown in Table 1 were obtained, with the true valuebeing: I (f ) = 1.0900. You are required to analyse the effect on the error of

n 4 8 16 32QS 0.866981048990 0.866973495703 0.866973019128 0.866972989326Error:

|QS − I (f )| × 1.0e05 0.8062 0.0508 0.0032 0.0002

Table 1: Composite Simpson rule: QS

halving the step size as shown. By use of these results show what you believeis the order of the composite Trapezoid rule. Show your results that lead youto this conclusion.

TURNOVER

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(d) When the integrand was replaced by f (x) = x2/3 the following results in Table2 were obtained, with the true value being: I (f ) = 2/5.

n 4 8 16 32QS 0.4000772494473 0.4000137134694 0.4000024278456 0.4000004294134Error:|QS − I (f )| × 1.0e04 0.7725 0.1371 0.0243 0.0043

Table 2: Composite Simpson rule: QS

Analyse the effect on the error of halving the step size. Explain your resultsand contrast them with your results from the above part (c). Explain why youthink this is happening.

TURNOVER

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4. (a) Consider the initial value problem (IVP)

dy

dx = f (x, y), y(0) = y0.

Then a 2 stage Runge-Kutta method for this IVP is

k1 = f (x, y)

k2 = f (x + h

2, y +

h

2k1)

yn+1 = yn + hk2

You are to use this RK-2 method to calculate an approximation to y(t) for thesecond order differential equation

d2ydt2

− y2 dydt

− y3 = t,

with initial conditions y(0) = 1, dydt

(0) = 3.

i. First write this equation as a matrix system of first order differential equa-tions showing all working.

TURNOVER

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ii. Write down the initial conditions for your system that make this probleman initial value problem.

iii. With step size h = 0.1 find an estimate for (y(0.1), y(0.1)) with the RK-2 method from part(a). Show all working, including the values for thevectors k1, and k2.

TURNOVER

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(b) An embedded tableau Runge-Kutta pair of respective orders 2 and 3 is usedto solve a first order initial value problem (IVP). This RK2(3) method with astep size h = 0.1 yields estimates for y(1.4) of yi for the 2rd order method, and

yi for the 3rd order method. These values for the given IVP are

yi = 1.58970 yi = 1.58981

i. Use these values to estimate the local error for the 2nd order method.Then, if the required accuracy of the LTE is ∆ = 10−4 decide whether thish step is acceptable.

ii. Now use an appropriate formula to calculate the next trial step size whena safety factor of α = 0.9 is assumed. Show all working and explain anydecisions you make.

TURNOVER

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(c) Suppose 100 independent observations are taken from the density in (1), andyou find x̄ = 0.70 and s = 0.24. Use the Central Limit Theorem to find anapproximate 95% confidence interval for E (X ). Recall that Φ(1.96) ≈ 0.975.

TURNOVER

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Emth271 12 Exam