1. In class we derived a bound on the relative error in the k-digit chopping representation of y. Show that y fl (y) y k+1

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1 APPM/MATH 4650 Problem Set 1 Solutions This assignment is due by 4:00pm Wednesday, September 11th. You may either turn it in to me in class or in the box outside my office door (ECOT 35). Minimal credit will be given for incomplete solutions or solutions that do not provide details on how the solution is found. For problems that require you to implement a method, you may use code from our textbook website faires/numerical-analysis/programs/. You may be asked to produce output from your implementation, but you do not need to turn in the actual code. 1. In class we derived a bound on the relative error in the k-digit chopping representation of y. Show that y k+1 where fl (y) is the k-digit rounding representation of y. Solution: Let the infinte-precision representation of y be given by y 0.d 1 d d k d k+1 10 n. Then the k-digit rounding approximation of y is given by fl (y) 0.d 1 d ˆd k 10 n where ˆdk We want to bound the relative error given by { dk if d k+1 < 5 d k + 1 if d k+1 5 y 0.d 1 d d k d k+1 10 n 0.d 1 d ˆd k 10 n 0.d 1 d d k d k+1 10 n We consider the two cases for ˆd k separately. Case 1: d k+1 < 5 ˆd k d k y 0.d 1 d d k d k+1 10 n 0.d 1 d d k 10 n 0.d 1 d d k d k+1 10 n 0.d k+1 d k+ 10 n k 0.d 1 d d k d k+1 10 n 0.d k+1 d k+ 0.d 1 d d k d k+1 10 k Since we ve assumed that that d k+1 < 5 we have that the numerator is bounded above by 0.5. The denomentator is bounded from below by 0.1 since d 1 0. We have y 0.d k+1 d k+ 0.d 1 d d k d k+1 10 k < k k+1

2 Case : d k+1 5 ˆd k d k + 1 y 0.d 1 d d k d k+1 10 n 0.d 1 d (d k + 1) 10 n 0.d 1 d d k d k+1 10 n (0.d 1 d d k d k+1 10 n 0.d 1 d d k 10 n ) 1 10 n k 0.d 1 d d k d k+1 10 n 0.d k+1 d k+ 10 n k 1 10 n k 0.d 1 d d k d k+1 10 n 0.d k+1 d k+ 1 0.d 1 d d k d k+1 10 k Since we ve assumed that that d k+1 5 the numerator is largest (in absolute value) when d k+1 5 and d k+ d k+3 0. The denomentator again is bounded from below by 0.1. So we have y 0.d k+1 d k+ 1 0.d 1 d d k d k+1 10 k k k+1 Then, from the combination of Case 1 and Case we have the desired result.. Consider the quadratic equation ax + bx + c 0 with a 1, b 400, and c The usual quadratic formula is given by x b ± b 4ac. (a) Find the roots of the equation using exact arithmetic. Solution: Using exactish arithmetic we have x 1 x (b) Find the roots of the equation using 10-digit chopping arithmetic and compute their relative errors. Discuss your results. Solution: Using chopped arithmetic we have x 1 x which gives relative errors of rel err (x 1 ) rel err (x )

3 The abysmal relative error in x is due to the catastrophic cancellation that occurs when computing b + b 4ac. (c) Show that if the negative root is given by the positive root can be computed via x 1 b b 4ac, x c b + b 4ac. Solution: If the negative root is given by then the positive root is given by x 1 b b 4ac, x b + b 4ac. Multiplying the top and bottom by the conjugate of the numerator we have x b + ( b 4ac b ) b 4ac b b 4ac b ( b 4ac ) ( b ) b 4ac 4ac ( b ) b 4ac c b + b 4ac. You may also remember that the two roots of the quadratic equation always satisfy x 1 x c a which can be solved for x to obtain the desired result. (d) Find the roots of the equation using the new formula and 10-digit chopping arithmetic and compute their relative errors. Discuss your results. Solution: We really only need to recompute x using the new formula. We have x which has relative erorr

4 rel err (x ) We now get a much better approximation for x since the new formula does not involve the subtraction of two similar values. 3. In this problem you will use the Bisection Method to compute the positive root of f (x) x 5 to within an absolute error of 10 4 using [1, 3] as the initial bracket. (a) Determine an upper bound on the number of iterations necessary to achieve the result. Solution: The bound on the error in the n th iterate is given by e n We then solve the following expression for n: (b a) n (b a) n 10 4 (3 1) n n 1 4 (n 1) log 10 n Max Iterations 15 log 10 (b) Run your code until the desired tolerance is reached. Your stopping criterion should be based on the theoretical bound on the absolute error since, in practice, we would not know the exact solution. Produce a table displaying for each step in the iteration: the current iterate p n, the absolute error e n p n p, and a theoretical upper bound on the absolute error e n. Solution: p n p p n b n (c) Plot the true absolute error e n versus the bound on the absolute error indicated by the theory. [Hint: Use a semi-log plot for a better idea of what is going on.] Discuss the plot.

5 Solution: We can see from the two plots that the true absolute error is always less than the theoretical error bound. We also notice that the true error tends to bounce around a lot. In fact, we see that on the 13 th iteration the true error actually falls below the defined tolerance of 10 4 and then gets mildly worse in the next iteration. We would never know this in practice though since we do not have access to the true error. We must rely instead on the theoretical error bound bound true error Error Iteration

6 4. Let f (x) (x 1) 10, p 1, and p n 1+1/n. Show that f (p n ) < 10 3 whenever n > 1 but that p p n < 10 3 requires n > Discuss the implications of this in terms of stopping criteria for iterative algorithms. [Hint: it might be helpful to look at the plot of f (x).] Solution: ( f (p n ) ) 10 n 1 1 n 10 If n then f (p n ) 1/ < 10 3 p p n (1 1 1 ) 1 n n < 10 3 n > 10 3 Suppose we implemented an iterative algorithm which produced the sequence p n converging to p and considered using a stopping criterion based on the residual (i.e. the size of f (p n )) or the absolute eroror p p n p n p n 1. For this particular problem if we used a tolerance of ɛ 10 3 the residual stopping criterion would tell us we are close enough after n 1 iteration. But, we can see from analysis of p p n that we need around n 1000 iterations to get an absolute error near the tolerance. The takeaway here is that for some functions, the residual f (p n ) can lie to you about the accuracy of your solution.