Program Explanation Calculation of Row1: Let the function be f(n) = log n, In this case, now it is needed to determine the largest “n” such that log n ≤ 1000000. To solve this inequality, rewrite the inequality as 2 log n ≤ 2 ⋅ 1000000 or n ≤ 2 ⋅ 1000000 . 1 second: It is already given that 2 10 ≈ 10 3 , thus we have that 2 1000000 = 2 10 ⋅ 100000 = ( 2 10 ) 100000 ≈ ( 10 3 ) 100000 = 10 300000 Thus, the function f ( n ) = l o g n for 1 second is 10 300000 microseconds. 1 hour: Similar to second, hour is calculated as follows: 2 3600000000 = 2 3600 ⋅ 1000000 = ( 2 3600 ) 1000000 = 2 3.6 × 10 9 Thus, the function f ( n ) = l o g n for 1 hour is 2 3.6 × 10 9 microseconds. 1 month: Similar to hour, month is calculated as follows: 2 2592000000000 = 2 2592 ⋅ 1000000000 = ( 2 2592 ) 1000000000 = 2 2.6 × 10 12 Thus, the function f ( n ) = l o g n for 1 day is 2 2.6 × 10 12 microseconds. 1 century: Similar to day, century is calculated as follows: 2 3110400000000000 = 2 31104 ⋅ 100000000000 = ( 2 31104 ) 100000000000 = 2 3.12 × 10 15 Thus, the function f ( n ) = l o g n for 1 century is 2 3.12 × 10 15 microseconds. Calculation of Row 2: Let the function be f(n) = n, In this case, now it is needed to determine the largest “n” such that n ≤ 1000000. 1 second: Since the given function is “n”. Let us take the value of 1 second directly. 1 second = 10 00000 = 10 6 Thus, the function f ( n ) = n for 1 second is 10 6 microseconds. 1 hour: Since the given function is “n”, let us take the value of 1 hour directly. Similar to second, hour is calculated as follows: 1 h o u r = 3600000000 = 3.6 × 10 9 m i c r o sec o n d s Thus, the function f ( n ) = n for 1 hour is 3 .6 × 10 9 microseconds. 1 month: Since the given function is “n”, let us take the value of 1 month directly. Similar to hour, month is calculated as follows: 1 m o n t h = 2592000000000 ( a s s u m e a m o n t h h a s 30 d a y s ) = 2.592 × 10 12 m i c r o sec o n d s Thus, the function f ( n ) = n for 1 day is 2.6 × 10 12 microseconds. 1 century: Since the given function is “n”, let us take the value of 1 second directly. Similar to day, century is calculated as follows: 1 c e n t u r y = 3110400000000000 = 3.1104 × 10 15 m i c r o sec o n d s Thus, the function f ( n ) = n for 1 century is 3.12 × 10 15 microseconds. Calculation of Row 3: Let the function be f(n) = n log n, let us use Maple to solve equations like n log n−1000000 = 0. The Maple command for solving this equation is, f s o l v e ( n × l o g [ 2 ] ( n ) − 1000000 = 0 ) In this case, now it is needed to determine the largest “n” such that n ≤ 1000000. 1 second: Since the given function is “n log n”, let us use the maple command specified above. From the given maple commands the equation is written as follows: n × l o g [ 2 ] ( n ) − 1000000 = 0 n × l o g [ 2 ] ( n ) = 10 6 n ( l o g [ 2 ] ) = 10 6 n = 10 6 l o g [ 2 ] n = 10 6 − l o g [ 2 ] ( log a b = log a − l o g b ) n = 10 6 − 10 1 n = 10 5 Thus, the function f ( n ) = n log n for 1 second is 10 5 microseconds. 1 hour: Since the given function is “n log n”, let us use the maple command specified above. From the given maple commands the equation is written as follows: n × l o g [ 2 ] ( n ) − 1000000000 = 0 n × l o g [ 2 ] ( n ) = 10 10 n ( l o g [ 2 ] ) = 10 10 n = 10 10 l o g [ 2 ] n = 10 10 − log [ 2 ] ( N o t e : log a b = log a − l o g b ) n = 10 10 − 10 1 n = 10 9 Thus, the function f ( n ) = n log n for 1 hour is 10 9 microseconds. 1 month: Since the given function is “n log n”, let us use the maple command specified above. From the given maple commands the equation is written as follows: n × l o g [ 2 ] ( n ) − 1000000000000 = 0 n × l o g [ 2 ] ( n ) = 10 12 n ( l o g [ 2 ] ) = 10 12 n = 10 12 l o g [ 2 ] n = 10 12 − l o g [ 2 ] ( N o t e : log a b = log a − l o g b ) n = 10 12 − 10 1 n = 10 11 Thus, the function f ( n ) = n log n for 1 day is 10 11 microseconds. 1 century: Since the given function is “n log n”, let us use the maple command specified above. From the given maple commands the equation is written as follows: n × l o g [ 2 ] ( n ) − 1000000000000000 = 0 n × l o g [ 2 ] ( n ) = 10 15 n ( l o g [ 2 ] ) = 10 15 n = 10 15 l o g [ 2 ] n = 10 15 − l o g [ 2 ] ( N o t e : log a b = log a − l o g b ) n = 10 15 − 10 1 n = 10 14 Thus, the function f ( n ) = n log n for 1 century is 10 14 microseconds

6th Edition

ISBN: 9781119278023

Author: Michael T. Goodrich; Roberto Tamassia; Michael H. Goldwasser

Publisher: Wiley Global Education US

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