# We need an upper bound for f 00 x on 1 2 1 noting

• Notes
• MohsinM1211
• 67
• 100% (2) 2 out of 2 people found this document helpful

This preview shows page 25 - 31 out of 67 pages.

We need an upper bound for f 00 ( x ) on 1 2 , 1 . Noting that x 3 is increasing on 1 2 , 1 = 2 x 3 is decreasing on 1 2 , 1 and positive throughout, thus | f 00 ( x ) | = 2 | x - 3 | ≤ 2 1 2 - 3 = 2 (2 - 1 ) - 3 = 2(8) = 16 = M. So using the error bound formula | f ( x ) - P 1 ( x ) | ≤ M 8 ( x 1 - x 0 ) 2 = 16 8 1 - 1 2 2 = 2 1 4 = 1 2 .
#6 Repeat the above for f ( x ) = x 1 3 , using the interval 1 8 , 1 .
25
Now f 0 ( x ) = 1 3 x - 2 3 and f 00 ( x ) = - 2 9 x - 5 3 = - 2 9 x 5 3 . Where is | f 00 ( x ) | a maximum on 1 8 , 1 ? f 00 ( x ) is monotonic (doesn’t change sign) on 1 8 , 1 , so check value at endpoints: f 00 1 8 = - 7 . ˙ 1 and f 00 (1) = - 0 . ˙ 2 = ⇒ | f 00 ( x ) | ≤ 7 . ˙ 1 = M . Thus from the error bound formula | f ( x ) - P 1 ( x ) | ≤ M 8 ( x 1 - x 0 ) 2 = 7 . ˙ 1 8 1 - 1 8 2 = 0 . 6806 (4 significant figures) 2.5 Trapezoid Rule Page 70 #1 Apply the trapezoid rule with h = 1 8 , to approximate the integral I = Z 1 0 1 1 + x 4 dx = 0 . 92703733865069 . How small does h have to be to get that the error is less that 10 - 3 ? 10 - 6 ?
26
i x i f ( x i ) 0 0 1 1 1 8 0 . 99987 ... 2 1 4 0 . 99805 ... 3 3 8 0 . 99025 ... 4 1 2 0 . 97014 ... 5 5 8 0 . 93145 ... 6 3 4 0 . 87157 ... 7 7 8 0 . 79400 ... 8 1 0 . 70710 ... T 8 ( f ) = 1 2 h ( f ( x 0 ) + 2 f ( x 1 ) + ... + 2 f ( x 7 ) + f ( x 8 )) = 1 2 1 8 (1 + 2(0 . 99987 ... ) + 2(0 . 99805 ... ) + ... + 2(0 . 79400 ... ) + 0 . 70710 ... ) = 0 . 926115180158 ... In order to estimate the error we need the 2 nd derivative of f ( x ) = (1+ x 4 ) - 1 2 . f 0 ( x ) = - 1 2 (1 + x 4 ) - 3 2 · 4 x 3 = - 2 x 3 (1 + x 4 ) - 3 2 . Using the product rule we can show that f 00 ( x ) = 6 x 2 ( x 4 - 1) ( x 4 + 1) 5 2 . This is a complicated function to get an upper bound on. We would use a combination of calculus and curve sketching techniques to deduce the follow- ing diagram: 27
Alternatively, you could get an upper bound by maximizing and minimizing the numerator and denominator in f 00 respectively. (Note: I wouldn’t give you one as complicated as this in our tests!) Now recall from lecture notes that to get an error 10 - 3 we require | I ( f ) - T n ( f ) | ≤ b - a 12 h 2 M = h 2 12 (1 . 39275) 10 - 3 , (8) where M = max | f 00 ( x ) | for all x [ a, b ] (as indicated in the diagram). Rearranging, we require h 2 0 . 08616047 ... or h 0 . 0928227 ... But h must divide b - a = 1 , so as n = b - a h or h = b - a n = 1 n 0 . 0928227 ... = n 10 . 77 The book would stop here. 28
so choose n = 11 = h = 1 n = 0 . 09. As in (8) we seek h such that h 2 12 (1 . 39275) 10 - 6 or h 2 8 . 616047 × 10 - 6 so h 0 . 0029353104 . As before h = 1 n 0 . 0029353104 ... = n 340 . 679 ... so with n = 341, = h = 1 n = 0 . 00293255132 ... #2 Use the trapezoid rule with h = π 4 , to approximate the integral I = Z π 2 0 sin( x ) dx = 1 . How small does h have to be to get that the error is less that 10 - 3 ? 10 - 6 ?
29
Thus | f 00 ( x ) | = | sin( x ) | ≤ 1 = M on h 0 , π 2 i . From the error bound formula we need | I ( f ) - T n ( f ) | ≤ b - a 12 h 2 M = π 2 12 h 2 · 1 = π 24 h 2 10 - 3 (9) = h 2 0 . 007639437 ... so h 0 . 08740387 ...