MATH ASSIGNMENT 03 SOLUTIONS

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1 MATH444.0 ASSIGNMENT 03 SOLUTIONS 4.3 Newton s method can be used to compute reciprocals, without division. To compute /R, let fx) = x R so that fx) = 0 when x = /R. Write down the Newton iteration for this problem and compute by hand or with a calculator) the first few Newton iterates for approximating /3 starting with x 0 = 0.5 and not using any division. What happens if you start with x 0 =? For positive R, use the theory of fixed point iteration to determine an interval about /R from which Newton s method will converge to /R. Solution. Given fx) = x R, we have f x) = x 2. The Newton iteration for this problem then is x k = x k fx k ) f x k ) = x k x k R = 2x k Rx 2 k. x 2 k To approximate /3, we use the above with R = 3. With x 0 = 0.5 we get x = 20.5) 30.5) 2 = 0.25 x 2 = 20.25) 30.25) 2 = x 3 = ) ) 2 = x 4 = ) ) 2 = which demonstrates rapid convergence to 0.3. On the other hand, with x 0 =, we see x = 2) 3) 2 = x 2 = 2 ) 3 ) 2 = 5 x 3 = 2 5) 3 5) 2 = 85 x 4 = 2 85) 3 85) 2 = 2845 which demonstrates rapid divergence. Considering Newton s method as a fixed-point iteration, we see that it will converge for x 0 chosen in an interval [a, b] which the function gx) = 2x Rx 2 maps into itself and for which g exists and is strictly bounded above by. Noting g is a polynomial and therefore differentiable, we note that g x) = 2 2Rx, and so g x) < 2R x 3 2R. All that remains to show is that g maps this interval into itself. To do this, maximize and minimize g on the interval [/2R, 3/2R]. The maximum occurs at /R and is /R while the minimum occurs at either endpoint, and takes the value 3/4R. Clearly g maps the interval into itself, and therfore for any choice of initial guess x 0 in [/2R, 3/2R] will give convergence.

2 4.2 Let ϕx) = x 2 + 4)/5. a) Find the fixed points) of ϕx). Solution. The fixed points of ϕ satisfy x = x and are the roots of fx) = x 2 5x + 4. The fixed points of ϕ are x =, 4. b) Would the fixed point iteration x k+ = ϕx k ) converge to a fixed point in the interval [0, 2] for all initial guesses x 0 [0, 2]? Solution. In order for fixed-point iteration to converge for all initial guesses x 0 [0, 2], ϕ must map [0, 2] into itself and ϕ x) < for all x [0, 2]. Note that ϕ x) = 2x. On [0, 2], 0 5 ϕ x) 4. Also, the minimum value of ϕx) 5 on this interval is 4 0 and the maximum value is 8 2. Thus ϕ satisfies the 5 5 hypotheses of the fixed-point theorem, and so fixed-point iteration converges to x = for all initial guesses in [0, 2]. 4.6 Steffensen s method for solving fx) = 0 is defined by where x k+ = x k fx k) g k, g k = f x k + fx k )) fx k ). fx k ) Show that this is quadratically convergent, under suitable hypotheses. Solution. Let gx) = fx + fx)) fx). fx) Supposing that x is a simple root of f that is f x ) 0), then gx), while not defined at x satisfies the property fx + fx)) fx) lim gx) = lim x x x x fx) = lim x x = f x ). f x + fx)) + f x)) f x) f x) Similarly, g is differentiable in a neighborhood of x, again, presuming that f is, and so we can argue that given φx) = x fx) gx) then φx) φx ) lim = 0 x x x x and therefore, Steffensen s method converges quadratically to the root x. 5.2 Write down the IEEE double-precision representation for the decimal number 50.2, using round to nearest. 2

3 Solution. Since 50.2 is positive, the sign bit here is s = 0. Also, 2 5 = < 2 6 = 64 so, the exponent is 5. Adding the bias, then, the exponent bits will be the bits that represent 028 = = 0x404, the last number being the hexadecimal representation. All that remains is to determine the mantissa. To do so, we need to note that 50 = = Also, we need to determine the binary representation of 0.2. Note that 0.2 = 5 = Here = 80 6 So far, then, we have = , so 5 80 = ) 80 = = ) Continuing, then it is easy to see that And so 0.2 = ) = ) = ) = ) ) Represented in binary, then, we have 50.2 = ) ). Rounding to nearest causes the last four bits to change from 00 2 to 00 2 because the bit after the fifty-second bit is that is, that 2 53 appears there). Putting all of the above together, we see that the double-precision representation of 50.2 in round to nearest is where the bars show the separation between the sign bit, the exponent and the mantissa. In hexadecimal this is written as 0x a. 5.3 What is the gap between 2 and the next larger double-precision number? 3

4 Solution. To determine the gap between 2 and the next larger number, determine the exponent for 2, and multiply by ε = Here 2 2 < 2 2, therefore the gap between 2 and the next larger double-precision number is 2 5. This can be verified using Matlab: log2eps2)) What is the gap between 20 and the next larger double-precision number? Solution. To determine the gap between 20 and the next larger number, determine the exponent for 2, and multiply by ε = Here 2 7 = < 2 8, therefore the gap between 20 and the next larger double-precision number is = This can be verified using Matlab: log2eps20)) -45 3) a) Show that ln x ) x 2 = ln x + ) x 2 Solution. Note that ) + ln ln x x 2 Therefore x + x 2 ) = ln x ) x 2 x + )) x 2 = ln x 2 x 2 ) ) = ln) = 0 ln x ) x 2 = ln x + ) x 2 b) Which of the two formulas is more suitable for numerical computation? Explain why and provide a numerical example in which the difference in accuracy is evident. Solution. The second formula, that is ln x + x 2 ), is more suitable for numerical computing because it doesn t suffer the cancellation that the first formula does. The following Matlab script demonstrates that the first formula is more inaccurate in the worst case than the second formula: f logx-sqrtx.ˆ2-)); % First formula g -logx+sqrtx.ˆ2-)); % Second formula 4

5 % Note that with x = sqrtyˆ2 + ), then expfx)) == x-y) should % be the case. We can use this to evaluate the accuracy in general % of the two formulae. Let's look at err x, y) absexpfx))-x-y))./absx))./eps; % which gives the relative error in terms of unit round-off. y = 0:0000; x = sqrty.ˆ2+); figure; plotx, errf, x, y)); hold on; plotx, errg, x, y)); setgca, 'XLim', [0 ceilmaxx))]); legend'error in First formula', 'Error in Second formula'); The generated plot looks like this: The other trouble is that the first formula results in for all double precision numbers larger than 2 53 while the second formula delays overflow until after realmax Here is an example: >> x = 2.^27:5); >> allisinffx))) >> allisfinitegx))) 5

6 4) For the following expressions, state the numerical difficulties that may occur, and rewrite the formulas in a way that is more suitable for numerical computation: a) x + x x, where x >>. x Solution. When x is much larger than, x 0 and so x + x x x x x, demonstrating catastrophic cancellation. This should be rewritten as x + x x x = x + x + x x x + x + x x ) ) x + x x x = x + x + x x 2 = ). x x + x + x x This formulation no longer suffers cancellation when x >> and will underflow more gradually as x. b) +, where a 0 and b. a 2 b 2 Solution. For a 0 the value will overflow whenever a < a 2 rmax where r max = realmax, the largest finite floating point number in a given floating-point system. As a result, getting a 2 out of the denominator is desired. One possible way to do this is a + 2 b = 2 a + a2 2 a 2 b 2 = a ) 2. + a b In the case that b < the above should be preferred and it won t suffer overflow until a becomes subnormal. If b >, b 2 +a 2 might be slightly preferred since ab that will delay overflow. c) a 2 + b 2 2ab sinθ) where a b and θ π 2. Solution. When a b and θ π 2, a2 + b 2 2ab sinθ) 0 and in fact, in floating point, may in fact turn out to be negative, which will never occur in exact arithmetic. For example, >> a = 0.3; b = a-epsa)/2; theta = pi/2); >> a.^2 + b.^2-2*a.*b.*sintheta) e-7 6

7 To maintain the property that this expression is never negative, rewrite it as a 2 + b 2 2ab sinθ) = a 2 sin 2 θ) + cos 2 θ) ) + b 2 2ab sinθ) = a 2 sin 2 θ) 2ab sinθ) + b 2 + a 2 cos 2 θ) = a sinθ) b) 2 + a cosθ)) 2 This is the sum of non-negative numbers and it will always be non-negative. 5) Consider the linear system a ) b x = b a) y 0) with a, b > 0; a b. a) If a b, what is the numerical difficulty in solving this linear system? Solution. It is easy to see that the correct solution to this linear system is a x = y = b a 2 b 2 a 2 b. 2 When a b the denominator suffers cancellation and becomes very small, resulting in inaccuracies when computing x, y. a b results in a nearly singular system, and in fact when a = b the system is singular and there is no solution. b) Suggest a numerically stable formula for computing z = x + y given a and b. Solution. Noting that a + b)x + a + b)y = then the sum z = x + y can be computed simply as z =. As long as a, b are a+b both positive and not subnormal, then z will be easily computed as there is no cancellation. c) Determine whether the following statement is true or false, and explain why: When a b, the problem of solving the linear system is ill-conditioned but the problem of computing x + y is not ill-conditioned. Solution. This statement is true. When a b, it is hard to compute x, y accurately because of cancellation in forming a 2 b 2, and the closer a is to b, the worse it gets. As demonstrated inn part b), however, when both a, b are positive, computing x + y can be easily done relatively accurately, regardless of the values of a, b, provided they are not too small. 7