MATH 104A HW 02 SOLUTION KEY

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1 .3 Algorithms and Convergence Solutions by Jon Lo Kim Lin - Fall 04 MATH 04A HW 0 SOLUTION KEY You are encouraged to collaborate with your classmates and utilize internet resources provided you adhere to the following rules: (i) You read the problem carefully and make a serious effort to solve it before you ask for help. (ii) You write the solution in your own words/ handwriting/ L A TEX document. (iii) You credit the people/books/websites who helped you, especially if they showed you how to do most of the problem. (iv) The TA is not omniscient. This course is aimed at developing your deductive reasoning and incredulity. For those who need a quick primer on programming, I highly recommend the python course by It s free and they claim you can learn python in a mere 3 hours..3 Algorithms and Convergence. (a) Use three-digit chopping arithmetic to compute the sum 0 i= (/i ) first by and then by Which method is more accurate, and why? (b) Write an algorithm to sum the finite series N i= x i in reverse order. SOLUTION. (a) Using three-digit chopping arithmetic, = (... (( ) + 0.) +... ) ) =.53, while = (... (( ) ) +... ) +.00) =.54. The actual value is The first sum is less accurate because the smaller numbers are added last, resulting in significant round-off error. (b) Without implementing three-digit chopping arithmetic, the pseudocode is given by Algorithm Calculate i=n x i INPUT: N, x,..., x N OUTPUT: i=n x i : i = N, Σ = 0. (initialize index and sum) : while i > 0 do 3: Σ = Σ + /i. (advance the sum) 4: i = i. (decrement the index) 5: end while 6: return sum.

2 .3 Algorithms and Convergence Solutions by Jon Lo Kim Lin - Fall 04 We now test the pseudocode to calculate actual value presented in part (a). The python code is listed below from math import f a c t o r i a l # Only i m p o r t what i s n e c e s s a r y # I n i t i a l i z e t h e i n d e x and sum i =0 sum =0.0 w h i l e i >0: sum=sum +.0/( i ) ; # I n c r e m e n t t h e sum i = i # Decrement t h e i n d e x p r i n t sum. The number e is defined by e = n=0 (/n!), where n! = n(n ) for n 0 and 0! =. Use four-digit chopping arithmetic to compute the following approximations to e, and determine the absolute and relative errors. (a) e 5 n=0 n! (c) e 0 n=0 n! (d) e 0 j=0 (0 j)! 5. Another formula for computing π can be deduced from the identity π/4 = 4 arctan 5 arctan 39. Determine the number of terms that must be summed to ensure an approximation to π to within 0 3. SOLUTION. A direct calculation makes it plain that the Maclaurin series for arctan x are given by x x x O(x 6 ), in the language of formal series we obtain: Using the identity: we obtain that π 4 = 4 = 4 k=0 arctan(x) = p k=0 k=0 ( ) k x k + k π 4 = 4 arctan 5 arctan 39, ( ) k ( 5 ) k + k ( ) k ( 5 ) k + k m=0 q m=0 for x <. ( ) m ( 39 ) m + m ( ) m ( 39 ) m + m + Error(p, q) Since the magnitude of the first term dominates the second term, it remains to identify the truncation number p for which the first sum is less than 0 3. Indeed, a direct calculation

3 .3 Algorithms and Convergence Solutions by Jon Lo Kim Lin - Fall 04 makes it plain that p = 3 is the smallest positive integer that suffices. The computation is routine and we ll spare you the details. 6. Find the rates of convergence of the following sequences as n. (a) (c) lim sin n n lim n = 0 (b) lim n sin n = 0 ( sin ) = 0 (d) lim [ln(n + ) ln(n)] = 0 n n SOLUTION. We first establish a seemingly "apparent" result to tackle parts (a) through (c). That is: 6 x sin(x) 4 π π 0 π π 4 6 To proof this rigorously, put f (x) = sin x x. On the interval (0, ) we see that f (x) = cos(x). Since cos(x) < it follows that f < 0 for all x. By invoking the mean value theorem we see that f is non-increasing on the interval (0, ). Since f is initially 0, i.e., f (0) = 0, it follows that f must be non-positive on the interval (0, ). That is, sin x x 0 on 0 < x <, in other words, sin x x, 0 < x <. () (a) Since 0 < n < for all n Z+, by the result established in () we deduce that sin n < n, () and it is now immediate that sin n O ( n). (b) Since 0 < n < for all n Z +, we obtain: sin n < n, therefore; sin n O ( n ). 3

4 .3 Algorithms and Convergence Solutions by Jon Lo Kim Lin - Fall 04 (c) By squaring both sides of () we see that sin < n, hence; it follows that ( ) sin n n. (d) Using the identity ln(a) ln(b) = ln( a b ), we see that ( ) n + ln(n + ) ln(n) = ln n Again, by graphing we find: n = ln( + n ). 6 x ln( + x) 4 π π 0 π π 4 6 Now, put g(x) = ln( + x). On the interval (0, ) we see that g (x) = x+. As g (x) < 0 on (0, ), again, by the mean value theorem the original function g is nonincreasing on the interval (0, ), and since g is initially zero, it must be the case that g is non-positive on the interval (0, ). In other words, ln( + x) < x for 0 < x <. Since 0 < n < it follows that ln(n + ) ln(n) = ln( + n ) < n, thus; ln(n + ) ln(n) O ( n). 8. (a) How many multiplication and additions are required to determine a sum of the form n i= i a i b j? (b) Modify the sum in part (a) to an equivalent form that reduces the number of computations. j= 4

5 .3 Algorithms and Convergence Solutions by Jon Lo Kim Lin - Fall 04 SOLUTION. First, recall the triangular number theorem, that is, Theorem. For every positive integer n, ( ) n n = = n(n + )/. For a rigorous argument see proof of the triangular number theorem. (a) Using the same technique presented in the first proof of the triangular number theorem, we proceed to represent our double summation in triangular form and see that n i= i a i b j = a b j= term + a b + a b -many terms + a 3 b + a 3 b + a 3 b 3 3-many terms. a n b + a n b + + a n b n + a n b n n-many terms, hence; in total, there are n = n(n + )/ many terms. Therefore, there are precisely n(n + )/ multiplications and n(n + )/ = (n + )(n ) many additions. (b) In the triangular representation, we notice the common factor a i in every term on the i-th row. By factoring and first summing over the j indices we see that n i= i a i b j = j= ( n n ) a i b j = j= i= n a i B i. Indeed, instead of i-many multiplications in the i-th row, we now require only one, namely that of a i and B i = b + b i.. Construct an algorithm that has as input an integer n, numbers x 0, x,..., x n, and a number x and that produces as output the product (x x 0 )(x x ) (x x n ). SOLUTION. Algorithm Calculate N i=0 (x x i) = (x x 0 )(x x ) (x x n ) INPUT: N, x = (x 0, x,..., x N ), x OUTPUT: N i=0 (x x i) : i = 0, p = (initialize index and product) : while i N do 3: p = p(x x i ) (advance the product) 4: i = i + (increment the index) 5: end while 6: return p i= 5

6 .3 Algorithms and Convergence Solutions by Jon Lo Kim Lin - Fall 04. Assume that 3 x x 4x + x + x x + x + 4x 3 8x 7 4 x 4 + x + = + x 8 + x + x, for x <, and let x = 0.5. Write and execute an algorithm that determines the number of terms needed on the left side of the equation so that the left side differs from the right side by less than 0 6. SOLUTION. Algorithm 3 Calculate INPUT: x, T OL, M OUTPUT: N DEFINITIONS: x M T OL N x x+x + x 4x 3 x +x + 4x 3 8x 7 4 x 4 +x + = +x 8 +x+x within error tolerance 0 6. = The point of evaluation = Maximum number of iterations = The error tolerance = The truncation number required to realize the desired error tolerance : sum = ( x)/( x + x ); : rhs = ( + x)/( + x + x ); 3: N = (initialize index) 4: while N M do 5: j = N 6: y = x j 7: a = jy x ( y) 8: b = y(y ) + 9: sum = sum + a/b (advance the sum) 0: if sum rhs < T OL then : return N : EXIT 3: else 4: N = N + (increment the truncation number) 5: end if 6: end while 7: print "Method failed. Please increase M" 3. (a) Suppose that 0 < q < p and that α n = α + O(n p ). Show that α n = α + O(n q ). 6

7 .3 Algorithms and Convergence Solutions by Jon Lo Kim Lin - Fall 04 (b) Make a table listing /n, /n, /n 3, and /n 4 for n = 5, 0, 00, and 000, and discuss the varying rates of convergence of these sequences as n becomes large. SOLUTION. First, fix n Z + and put f (x) = n x. By differentiating implicitly we find that f (x) = f (x) ln n. By the familiar argument, that is, by way of the mean value theorem, we deduce that f is non-decreasing on R +. (a) Since α n α O(n p ), it follows by definition that for sufficiently large n there exists some constant K for which α n α K n p. (3) Since 0 < q < p, we see that n p < n q for any n Z +. Using the same K from (3) it is now immediate that α n α K n p K n q, so α n α O(n q ) as required. (b) From the tableau below we see that the most rapid convergence is O(/n 4 ). n /n /n /n 3 /n e e e-03.60e e-0.00e-0.00e-03.00e e-0.00e-04.00e-06.00e e-03.00e-06.00e-09.00e- The corresponding python code is posted below: # Compare t h e v a r y i n g r a t e s o f c o n v e r g e n c e f o r n i n [ 5. 0, 0. 0, , ] : p r i n t "n=%.0 f / n=%. e / n^=%. e " \ "/ n^3=%. e / n^4=%. e " % \ ( n, /n, /( n ), /( n 3 ), /( n 4 ) ) 5. Suppose that as x approaches zero, F (x) = L + O(x α ) and F (x) = L + O(x β ). Let c and c be nonzero constants, and define F (x) = c F (x) + c F (x) G(x) = F (c x) + F (c x). and Show that if γ = min{α, β}, then as x approaches zero, (a) F (x) = c L + c L + O(x γ ) (b) G(x) = L + L + O(x γ ). 7

8 .3 Algorithms and Convergence Solutions by Jon Lo Kim Lin - Fall 04 SOLUTION. Suppose that for sufficiently small x we have positive constants K, and K independent of our choice of x, for which the following two inequalities hold simultaneously, F (x) L K x α and F (x) L K x β. Now, choose c = max{ c, c, }, K = max{k, K }, and µ = max{α, β}. (a) A direct calculation using the Triangle inequality makes it plain that for sufficiently small x, F (x) c L c L = c (F (x) L ) + c (F (x) L ) c (F (x) L ) + c (F (x) L ) = c (F (x) L ) + c (F (x) L ) c (F (x) L ) + c (F (x) L ) ck x α + ck x β ck x α + ck x β = ck ( x α + x β) ck x γ ( + x µ γ) ck x γ. By putting K = ck, it is now immediate that F (x) c L c L O(x γ ), i.e., as required. F (x) = c L + c L + O(x γ ). (b) Again, for sufficiently small x, the Triangle inequality makes it plain that G(x) L L = F (c x) + F (x x) L L (4) K c x α + K c x β (5) Kc µ ( x α + x β) (6) Kc µ x γ ( + x µ γ) (7) Kc µ x γ. (8) By putting K = Kc µ, it is now immediate that G(x) L L O(x γ ), i.e., G(x) = L + L + O(x γ ). 6. The sequence {F n } described by F 0 =, F =, and F n+ = F n + F n+, if n 0, is called a Fibonacci sequence. Its terms occur naturally in many botanical species, particularly those with petals or scales arranged in the form of a logarithmic spiral. Consider the sequence {x n }, where x n = F n+ /F n. Assuming that lim n x n = x exists, show that x = ( + 5 ) /. This number is called the golden ratio. We highly recommend that you watch the YouTube video by Vi Hart 8

9 .3 Algorithms and Convergence Solutions by Jon Lo Kim Lin - Fall 04 SOLUTION. Give that x n x as n we see that also, lim x n+ = x = lim x n, n n x n+ = F n+ F n = F n + F n+ F n = F n F n + F n+ F n = + x n, hence; letting n on both sides of our equality, we obtain: In other words, x = + x. x x = 0, whose roots, by means of the quadratic formula, are precisely: x = ( ± 5)/. We choose the positive root as F n > 0 by construction. The python code below generates the smallest Fibonacci number below 000, namely: 597. c l a s s F i b s : # C r e a t e t h e F i b o n a c c i c l a s s def init ( s e l f ) : s e l f. a = 0 s e l f. b = def n e x t ( s e l f ) : s e l f. a, s e l f. b = s e l f. b, s e l f. a+ s e l f. b r e t u r n s e l f. a def iter ( s e l f ) : r e t u r n s e l f f i b s = F i b s ( ) # Make a F i b s o b j e c t # f i n d t h e s m a l l e s t F i b o n a c c i number >=,000 f o r f i n f i b s : i f f > : p r i n t f break Although we did not ask you to prove the convergence of the sequence {x n } n=, we include the proof below for the sake of completeness. Don t worry if you cannot follow the arguments 9

10 .3 Algorithms and Convergence Solutions by Jon Lo Kim Lin - Fall 04 after a first reading as they employ proof techniques from MATH 8. Without further ado, by construction our sequence {x n } is non-decreasing, that is, x n x n+. Also, x n = f n+ f n = f n + f n f n = f n f n + f n f n = f n f n +. Since f n f n for all n, we see that f n /f n, hence; x n, indeed the sequence {x n } is bounded from above. Let E denote the range of the sequence {x n }. By the completeness axiom of the real line, R, our set E must admit a least upper bound (also known as supremum), say x 0. More specifically, for any ε > 0, there exists some positive integer N sufficiently large such that x 0 ε < x N x 0, for otherwise, x 0 ε would be an upper bound for E, thus, contradicting the position of x 0 as the least upper bound. Since our sequence {x n } is non-decreasing, it follows that x 0 ε < x n x 0, as soon as n N. This proves that x n x 0 as n and we establish the existence of a limit as desired. 0