Math 128A: Homework 2 Solutions

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1 Math 128A: Homework 2 Solutions Due: June In problems where high precision is not needed, the IEEE standard provides a specification for single precision numbers, which occupy 32 bits of storage. Look up this specification on the web, such as the number of bits attributed to the mantissa and exponent, the machine precision ɛ, the range of representable numbers, etc. Give the sequence of bits representing the real number in single precision. Single precision uses one bit for the sign s, 8 bits for the exponent and 23 bits for the fractional part of the mantissa. There are 2 8 = 256 possible exponent values; in order to represent numbers less than 1 as well as larger than one, we take the exponent to be e = c 127, where c is the number encoded by the 8 exponent bits. It follows that e { 127, 126,..., 127, 128}. Meanwhile, the mantissa yields M = m m m so the corresponding machine number is ( 1) s (1 + M) 2 e. Recall that machine epsilon ɛ is defined as the difference between 1 and the next machine number a. We have a = ( ) 2 0 so it follows that ɛ = The smallest representable positive number S corresponds to e = 126 and M = 0 so that S = (1 + 0) Likewise, the largest representable positive number L corresponds to e = 127 and M = = so that L = ( ) Finally, we turn to the single precision representation of Note first that (11.25) 10 = ( ) 2 = ( ) 2. The sign bit is 0 as the number is positive and the mantissa is As the exponent is 3, it corresponds to c = = 130 which, in binary, is To sum up, we have the representation Suppose that as x a, f 1 = O(g 1 ) and f 2 = O(g 2 ). Prove that 1

2 (a) f 1 + f 2 = O( g 1 + g 2 ). For x sufficiently close to a, we have f 1 (x) K 1 g 1 (x) and f 2 (x) K 2 g 2 (x) for some constants K 1, K 2. It follows that f 1 (x) + f 2 (x) f 1 (x) + f 2 (x) K 1 g 1 (x) + K 2 g 2 (x) K 3 ( g 1 (x) + g 2 (x) ) where K 3 = max(k 1, K 2 ). It follows that f 1 + f 2 = O( g 1 + g 2 ). (b) f 1 f 2 = O(g 1 g 2 ). For x sufficiently close to a, we have f 1 (x) K 1 g 1 (x) and f 2 (x) K 2 g 2 (x) for some constants K 1, K 2. It follows that f 1 (x)f 2 (x) = f 1 (x) f 2 (x) (K 1 g 1 (x) )(K 2 g 2 (x) ) = K 3 g 1 (x)g 2 (x) where K 3 = K 1 K 2. It follows that f 1 f 2 = O(g 1 g 2 ). (c) for any constant c, cf 1 = O(g 1 ). For x sufficiently close to a, we have f 1 (x) K 1 g 1 (x) so that whence it follows that cf 1 = O(g 1 ). cf 1 (x) = c f 1 (x) ( c K 1 ) g 1 (x) 3. In class, we saw that k-digit chopping is accurate to at least (k 1) significant figures. In this problem, we find the analogous result for k-digit rounding. (a) Convince yourself that, given a decimal number of the form a = d 1.d 2 d 3... d k d k n with d 1 0, k-digit rounding is equivalent to k-digit chopping applied to a n k. This equivalency is best illustrated by an example. Suppose k = 3, a 1 = and a 2 = Then, defining ã i = a i gives ã 1 = , ã 2 = whence 3-digit chopping yields â 1 = and â 2 = This is clearly in agreement with our understanding of rounding. (b) By combining the steps in the chopping proof with (a), show that k-digit rounding is accurate to at least k significant figures. Suppose â is the result of k-digit rounding applied to a. From (a), it follows that a â 5 10 n k so that a â a 5 10 n k d 1.d 2 d n k = 5 10 k 2

3 where we used the fact that d 1.d 2 d since d 1 0. We conclude that k-digit rounding is accurate to at least k significant figures. 4. Write a MATLAB function of the form function p = Horner(z,A) that uses Horner s method to evaluate the polynomial p(x) = a n x n + a n 1 x n a 1 x + a 0 at a point z with A = [a 0, a 1, a 2,..., a n 1, a n ]. 1. Define b n = a n. 2. Define b i 1 = a i 1 + zb i for i = n, n 1,..., Output b Use the MATLAB function BisMethod.m to solve the following equations and describe your results (that is, the interval you started from, the tolerance chosen, the number of iterations needed to get the desired accuracy and, finally, the answer). (a) x x = 50. Let f(x) = x x 50. Note that as f(3) = 23 < 0 and f(4) = 206 > 0 and f is continuous on [3, 4], by the IVT f(x) = 0 has a solution in p (3, 4). Running the bisection method with tol = yields p using 40 iterations. (b) ln(x) = cos(x). Let f(x) = ln(x) cos(x). Note that as f(1) = ln(1) cos(1) = cos(1) < 0 and f(π/2) = ln(π/2) cos(π/2) = ln(π/2) > 0 and f is continuous on [1, π/2], by the IVT f(x) = 0 has a solution in p (1, π/2). Running the bisection method with tol = yields p using 40 iterations. (c) x + e x2 = 0. Let f(x) = x + e x2. Note that as f( 1) = 1 + e 1 < 0 and f(0) = e 0 > 0 and f is continuous on [ 1, 0], by the IVT f(x) = 0 has a solution in p ( 1, 0). Running the bisection method with tol = yields p using 40 iterations. 6. An eclipse of the sun is said to be α-complete when α% of the area of the sun s disc is hidden behind the disc of the moon. One can model this with two discs, each of radius 1. Use the bisection method to find the distance between the centers of the discs when the eclipse is 10-complete. Let θ be the angle as shown in Figure 1. Then, the common area is ( 1 A(θ) = 2 2 (1)2 θ 1 ) 2 (1)2 sin(θ) = θ sin(θ). 3

4 We wish to solve A(θ) = απ/100 = π/10, i.e., θ sin(θ) = π/10. Let f(θ) = θ sin(θ) π/10. Then, f(0) = π/10 < 0 and f(π) = π π/10 > 0; as f is continuous everywhere, it follows from IVT that f(θ) = 0 has a solution θ 0 (0, π). Running the bisection method with tol = yields θ using 42 iterations. The distance between the centres is d = 2 cos(θ 0 /2) Figure 1 7. Transform the following equations algebraically into a fixed point form and compute solutions using the fixed point MATLAB solver FixPoi.m. (a) x ln(x) 1 = 0. We have x ln(x) 1 = 0 ln(x) = 1/x x = e 1/x. Using this form with p 0 = 1 and setting tol = leads to the solution p with 51 iterations. (b) x + x = 1 + x 2. We have x 2 = x + x 1 x = x x x = (x x) 2. Using this form with p 0 = 0.9 and setting tol = leads to the solution p with 20 iterations. (c) 3x 2 + tan(x) = 0. We have 3x 2 = tan(x) x = tan(x) 3x. Using this form with p 0 = 1.9 and setting tol = leads to the solution p with 13 iterations. 4

5 8. Let x = g(x) be a fixed point form for which the sequence {p n } diverges, because g (x) > 1. (a) Prove that then the fixed point form x = g 1 (x) generates a convergent sequence to the fixed point p of x = g(x). We assume first that g is continuous and differentiable on an interval I R and that it has a fixed point at p I. We also assume that it is invertible. Let the range of g be denoted by J; it follows that g 1 is continuous on J. Recall then that whenever (g 1 ) (x) exists, (g 1 ) 1 (x) = so that g (g 1 (x)) (g 1 ) (x) = 1 g (g 1 (x)) < 1 where we used the fact that g > 1. This also shows that g 1 is differentiable on the entire interval J. Finally, let δ > 0 be such that [p δ, p + δ] J. For any x [p δ, p + δ], we then have from the mean value theorem g 1 (x) g 1 (p) x p = (g 1 ) (c(x)) < 1 g 1 (x) p < x p which proves that the restriction of g 1 to [p δ, p + δ] maps into [p δ, p + δ]. It follows from the fixed point iteration convergence theorem that p n+1 = g 1 (p n ) with p 0 [p δ, p + δ], generates a sequence that converges to p. (b) Compute the first positive solution of x tan x = 0. Hint: The fixed point form x = tan(x) does not converge; however, using (a), it can be argued that x = arctan(x) + π generates a convergent sequence over an appropriate interval. Let g : (5π/4, 3π/2) (1, ) be defined by g(x) = tan(x). Note then that g is continuous and differentiable on (5π/4, 3π/2). It is also invertible with g 1 : (1, ) (5π/4, 3π/2) given by g 1 (x) = arctan(x) + π (where arctan( ) maps to ( π/2, π/2)). Note furthermore that g (x) = sec 2 (x) g (x) = sec 2 (x) > 1 as x (5π/4, 3π/2). It follows from (a) that the form x = g 1 (x) generates a convergent sequence. Using this form with p 0 = 4 and setting tol = leads to the solution p with 10 iterations. 9. (i) Determine analytically the fixed point that is computed by the following iterations. 5

6 ( (a) x 0 = 1, x n+1 = 0.2 4x n + a x n ). This iteration comes from the fixed point form ( x = g 1 (x) := 0.2 4x + a ) 5x = 4x + a x x x2 = a x = ±a 1/2. In addition, g 1(x) = 0.2 (4 a ) g x 1(a 1/2 ) = ( 1, 1) which shows that starting close to x = a 1/2, the sequence will converge to a 1/2. ( (b) x 0 = 1, x n+1 = 0.5 x n + a x n ). This iteration comes from the fixed point form ( x = g 2 (x) := 0.5 x + a ) 2x = x + a x x x2 = a x = ±a 1/2. In addition, g 2(x) = 0.5 (1 a ) g x 2(a 1/2 ) = 0 ( 1, 1) 2 which shows that starting close to x = a 1/2, the sequence will converge to a 1/2. (c) x 0 = 1, x n+1 = xn(x2 n +3a) 3x 2 n+a. This iteration comes from the fixed point form x = g 3 (x) := x(x2 + 3a) 3x 2 + a 3x 3 + ax = x 3 + 3ax x 3 ax = 0 x = 0, ±a 1/2. In addition, g 3(x) = 3(x4 2ax 2 + a 2 ) (3x 2 + a) 2 g 3(a 1/2 ) = 0 ( 1, 1) which shows that starting close to x = a 1/2, the sequence will converge to a 1/2. (ii) Examine, numerically, the speeds of convergence for a = 10 (i.e., determine the number of iterations needed to converge for all three and rank them in order of leastto-most iterations). Using tol = 10 12, (a) takes 52, (b) takes 6 and (c) takes 4 iterations. This happens because (a) is linear convergence with g (a 1/2 ) = 3/5 < 1, (b) is quadratic with g (a 1/2 ) = 0 and (c) is cubic convergence with g (a 1/2 ) = g (a 1/2 ) = 0. 6