Math128_hw5

# Math128_hw5 - f x 4 where C i = R b a L i x dx These...

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Math 128A, Spring 2007 Homework 5 Solution 5.2.8 (a) I ( f ) = R b a f ( x ) dx R b a f (( a + b ) / 2) dx = ( b - a ) f (( a + b ) / 2) = M 1 ( f ). (b) If we subdivide [ a,b ] into n subintervals a = x 0 < x 1 < ··· < x n = b with x i +1 = x i + h and apply M 1 to each subinterval, we get I ( f ) = Z b a f ( x ) dx = Z x 1 x 0 f ( x ) dx + ··· + Z x n x n - 1 f ( x ) dx ( x 1 - x 0 ) f (( x 0 + x 1 ) / 2) + ··· + ( x n - x n - 1 ) f (( x n - 1 + x n ) / 2) = hf ( a + h/ 2) + ··· + hf ( a + ( n - 1 / 2) h ) (c) M 1 (1 / (1 + x )) = 2 / 3 . 6667 and M 2 (1 / (1 + x )) = 2 / 5 + 2 / 7 . 6857, while I = log 2 . 6931. 5.2.10 (a) Following the derivation of Simpson’s rule, I ( f ) = Z b a f ( x ) dx Z b a P 4 ( x ) dx = Z b a [ f ( x 0 ) L 0 ( x ) + ··· + f ( x 4 ) L 4 ( x )] dx = C 0 f ( x 0 ) + ··· + C 4
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Unformatted text preview: f ( x 4 ) where C i = R b a L i ( x ) dx . These integrals are easily evaluated (but it is very tedious!), and you get Boole’s rule. (b) B 4 (1 / (1+ x )) = (2 / 45) * (1 / 4) * (7+32 * 4 / 5+13 * 2 / 3+32 * 4 / 7+7 * 1 / 2) ≈ . 7006. This is just a little worse than M 2 (1 / (1 + x )) from problem 8. 1...
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