ps1solution - 13.002 PS #1 Solution 1.2.4. (a) (1.0110101)2...

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13.002 PS #1 Solution 1.2.4. (a) (1.0110101) 2 = (2^0)+(2^-2)+(2^-3)+(2^-5)+(2^-7)= 1.41406250000000 1.2.4. (b) (11.0010010001) 2 = (2^1)+(2^0)+(2^-3)+(2^-6)+(2^-10)= 3.14160156250000 1.2.5. (a) 2 (1.0110101) = 1.41421356237309 1.41406250000000= 2 = 1.510623730900385e-004 (absolute error) 1.510623730900385e-004 = 1 .068172283940988e-004 (relative error) 1 .41421356237309 1.2.5. (b) π (11.0010010001) 2 = 3.14159265358979- 3.14160156250000 = -8.908910209992627e-006 (absolute error) 8.908910209992627e-006 = 2.835794194964366e-006 (relative error) 3.14159265358979 1.2.13. (b) 1 1 1 ( + ) + = 10 3 5 × 2 1 ((0.1101) × 2 3 + (0.1011) ) = (0.1110) × 2 1 2 2 2 × 2 2 (0.1110) × 2 1 + (0.1101) = (0.1010) × 2 0 2 2 2 19 (0.1010) × 2 0 30 2 0.63333333333333- 0.6250000000000= 0.008 3 333333333 (absolute error) 0.0083333333333 = 0.01315. .. (relative error) 0.63333333333333
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1.3.1.(b) 98350-98000 = 350 (absolute error) 350 = 0 .00355871886121 (relative error) (2 significant digits)
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This note was uploaded on 02/24/2012 for the course MECHANICAL 2.993J taught by Professor Henrikschmidt during the Spring '05 term at MIT.

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ps1solution - 13.002 PS #1 Solution 1.2.4. (a) (1.0110101)2...

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