(Floating point) Write 98.17, -100.987, 0.0057869, - 13600 in floating-point form, rounded to 4S (4 significant digits).
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Textbook Solutions for Advanced Engineering Mathematics
Question
TEAM PROJECT. Romberg Integration (W. Romberg. Norske Videllskab. Trolldheil11, FfJrh. 28, Nr. 7, 1955). This method uses the trapezoidal rule and gains precision stepwise by halving h and adding an error estimate. Do this for the integral of f(x) = e-X ti'om x = 0 to x = 2 with TOL = 10-3 , as follows. Step 1. Apply the trapezoidal rule (2) with h = 2 (hence n = I) to get an approximation In. Halve II and use (2) to gel 121 and an error estimate 1 E21 = 22 _ I (121 - 1u) If IE211 ~ TOL, stop. The result is 122 = 121 + E21' Step 2. Show that E21 = -0.066596, hence 1"211 > TOL and go on. Use (2) with hl4 to get 131 and add to it the error estimate 1"31 = i(131 - 1 21 ) to get the better 132 = 131 + 1"31- Calculate I I E32 = 24 _ I (132 - 1 22) = 15 (132 - 1 22),If IE3d ~ TOL, stop. The result is is3 = 132 + E32.(Why does 24 = 16 come in?) Show that we obtain1"32 = -0.000266, so that we can stop. Arrange your1- and E-values in a kind of "difference table."If IE3d were greater than TOL, you would have [0go on and calculate in the next step 141 from (1) withh = ~: thenI 142 = 141 + E41 with 1"41 = "3 (141 - 1 31)I 143 = 142 + 1"42 with q2 = 15 (142 - 1 32)I 144 = 143 + E43 with E43 = 63 (J43 - is3)where 63 = 26 - 1. (How does this come in?)Apply the Romberg method to the integral off( 1:) = ~7TX4 cos !7TX from x = 0 to 2 with TOL = 10-4 .
Solution
The first step in solving 19 problem number 26 trying to solve the problem we have to refer to the textbook question: TEAM PROJECT. Romberg Integration (W. Romberg. Norske Videllskab. Trolldheil11, FfJrh. 28, Nr. 7, 1955). This method uses the trapezoidal rule and gains precision stepwise by halving h and adding an error estimate. Do this for the integral of f(x) = e-X ti'om x = 0 to x = 2 with TOL = 10-3 , as follows. Step 1. Apply the trapezoidal rule (2) with h = 2 (hence n = I) to get an approximation In. Halve II and use (2) to gel 121 and an error estimate 1 E21 = 22 _ I (121 - 1u) If IE211 ~ TOL, stop. The result is 122 = 121 + E21' Step 2. Show that E21 = -0.066596, hence 1"211 > TOL and go on. Use (2) with hl4 to get 131 and add to it the error estimate 1"31 = i(131 - 1 21 ) to get the better 132 = 131 + 1"31- Calculate I I E32 = 24 _ I (132 - 1 22) = 15 (132 - 1 22),If IE3d ~ TOL, stop. The result is is3 = 132 + E32.(Why does 24 = 16 come in?) Show that we obtain1"32 = -0.000266, so that we can stop. Arrange your1- and E-values in a kind of "difference table."If IE3d were greater than TOL, you would have [0go on and calculate in the next step 141 from (1) withh = ~: thenI 142 = 141 + E41 with 1"41 = "3 (141 - 1 31)I 143 = 142 + 1"42 with q2 = 15 (142 - 1 32)I 144 = 143 + E43 with E43 = 63 (J43 - is3)where 63 = 26 - 1. (How does this come in?)Apply the Romberg method to the integral off( 1:) = ~7TX4 cos !7TX from x = 0 to 2 with TOL = 10-4 .
From the textbook chapter Numerics in General you will find a few key concepts needed to solve this.
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