10_Cal_Solution of Calculus_6e

10_Cal_Solution of Calculus_6e - in the form x = k + d 1 10...

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C01S01.053: By the result of Problem 52, the range of Round (10 x ) is the set of all integers, so the range of g ( x )= 1 10 Round (10 x ) is the set of all integral multiple of 1 10 . C01S01.054: What works for π will work for every real number; let Round 2( x )= 1 100 Round (100 x ). To be certain that this is correct (we will verify it only for positive numbers), write the [positive] real number x in the form x = k + t 10 + h 100 + m 1000 + r, where k is a nonnnegative integer, t (the “tenths” digit) is a nonnegative integer between 0 and 9, h (the “hundredths” digit) is a nonnegative integer between 0 and 9, as is m , and 0 5 r< 0 . 001. Then Round 2( x )= 1 100 Floor (100 x +0 . 5) = 1 100 Floor (100 k +10 t + h + 1 10 ( m +5)+100 r ) . If 0 5 m 5 4, the last expression becomes 1 100 (100 k +10 t + h )= k + t 10 + h 100 , which is the correct two-digit rounding of x .I f5 5 m 5 9, it becomes 1 100 (100 k +10 t + h +1)= k + t 10 + h +1 100 , also the correct two-digit rounding of x in this case. C01S01.055: Let Round 4( x )= 1 10000 Round (10000 x ). To verify that Round 4 has the desired property for [say] positive values of x , write such a number
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Unformatted text preview: in the form x = k + d 1 10 + d 2 100 + d 3 1000 + d 4 10000 + d 5 100 , 000 + r, where k is a nonnegative integer, each d i is an integer between 0 and 9, and 0 5 r < . 00001. Application of Round 4 to x then produces 1 10000 Floor (10000 k + 1000 d 1 + 100 d 2 + 10 d 3 + d 4 + 1 10 ( d 5 + 5) + 10000 r ) . Then consideration of the two cases 0 5 d 5 5 4 and 5 5 d 5 5 9 will show that Round 4 produces the correct four-place rounding of x in both cases. C01S01.056: Let Chop 4( x ) = 1 10000 Floor (10000 x ). Suppose that x > 0. Write x in the form x = k + d 1 10 + d 2 100 + d 3 1000 + d 4 10000 + r, where k is a nonnegative integer, each of the d i is an integer between 0 and 9, and 0 5 r < . 0001. Then Chop 4( x ) produces 1 10000 Floor (10000 k + 1000 d 1 + 100 d 2 + 10 d 3 + d 4 + 10000 r ) = 1 10000 (10000 k + 1000 d 1 + 100 d 2 + 10 d 3 + d 4 ) 10...
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This note was uploaded on 09/14/2011 for the course MATH 101 taught by Professor Cheng during the Spring '11 term at Ecole Hôtelière de Lausanne.

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