Unformatted text preview: accuracy is not enough to oﬀset the superior accuracy of the ﬁvepoint NewtonCotes rule on the original interval.
(c ) For an npoint interpolatory quadrature rule Qn , we have the general error
bound
1
E ≤ hn+1 f (n) ∞ .
4
The comparable bound for the rule Q2n−1 on an interval of twice the length is COT 4501 E≤ 1
(2h)2n f (2n−1) ∞ .
4 Solutions for Exam Review Problems Again, provided the relevant derivatives of f are well behaved, we see that the
(Chapter 8 and 9) higherdegree rule is more accurate than the lowerdegree rule applied twice on
intervals of half the size.
8.12. Averaging the ﬁrstorder accurate forward and backward diﬀerence formulas,
we obtain
f (x + h) − f (x − h)
1 f (x + h) − f (x) f (x) − f (x − h)
+
=
,
2
h
h
2h
which is the centered diﬀerence formula we already know to be secondorder accurate from the analysis in Section 8.6.1.
8.13. In the Taylor series expansion
f (x) 2 f (x) 3
f (x + h) = f (x) + f (x)h +
h+
h + ···,
2
6
we solve for f (x) to obtain
f (x) = f (x) 2
f (x + h) − f (x) f (x)
−
h−
h + ···.
h
2
6 Similarly, in the Taylor series expansion
f (x)
f (x)
(2h)2 +
(2h)3 + · · ·
2
6
4f (x) 3
= f (x) + 2f (x)h + 2f (x)h2 +
h + ···,
Chapter 8: Numerical Integration and Diﬀerentiation
3 f (x + 2h) = f (x) + f (x)(2h) + 104 we solve for f (x) to obtain
f (x) = 2f (x) 2
f (x + 2h) − f (x)
− f (x)h −
h + ···.
2h
3 If we now subtract the second of these series for f (x) from twice the ﬁrst, we obtain
f (x) = −3f (x) + 4f (x + h) − f (x + 2h) f (x) 2
+
h + ···,
2h
3 so that −3f (x) + 4f (x + h) − f (x + 2h)
2h
is a secondorder accurate, onesided diﬀerence approximation to f (x).
f (x) ≈ 8.14.
F (0) = F (h) + F (h) − F (h/2)
−0.8333 + 0.9091
= −0.8333 +
= −0.9849.
−1 − 1
2
0.5 − 1 8.15. (a ) The power series expansion for the sine function is 2h 3 so that −3f (x) + 4f (x + h) − f (x + 2h)
2h
is a secondorder accurate, onesided diﬀerence approximation to f (x).
f (x) ≈ 8.14.
F (0) = F (h) + F (h) − F (h/2)
−0.8333 + 0.9091
= −0.8333 +
= −0.9849.
−1 − 1
2
0.5 − 1 8.15. (a ) The power series expansion for the sine function is
sin(x) = x −
so
pn x5
x7
x3
+
−
+ ···,
3!
5!
7!
π
π3
π5
π7
π3
π5
π7
=n
− 3 + 5 − 7 + ··· = π − 2 + 4 − 6 + ···
n n 3! n 5! n 7!
n 3! n 5! n 7!
5
7
3
π2π4π6
h+
h−
h + ···,
= π−
3!
5!
7! where h = 1/n, and hence we have a0 = π . Similarly, the power series expansion
Exercises
for the tangent function is
x3
2x5
17x7
+
+
+ ···,
3
15
315
u = u2 ,
so
1
u π 7= u3 ,
2
π3
2π 5
17
π3
2π 5
17π 7
π
+3+5 +7
+ ··· = π + 2 + 4 + 6
+ ···
qn = n
u=
n n 3 n 15 n3 315 −u1 u3 .
n 3 n 15 n 315
3
5
7
(c ) = π + π h2 + 2π n4 + 17π n6 + · · · ,
3
15
315
u1 = u2 ,
and hence we have b0 = π. (b ) Using Richardson 2extrapolation with p = 2 and
u2 = −GM u1 /(u2 + u3 )3/2 ,
1
q = 2, we have
u3 = u4 ,
F (1/6) − F (1/12)
3.0000 − 3.1058
u−2 = −GM u3 /0000 + 2 )3/2 .
a0 = F (1/6) +
= 3. (u2 + u3
= 3.1411.
4
1
2 −1
0.25 − 1 (b ) tan(x) = x + 9.3. Yes, since thisF (1/6) − F (1/12) homogeneous4641 − 3.with constant coeﬃﬁrstorder, linear,
3. system 2154
b0 = F (1written
/6) +
= 3.4641 +
= 3.1325.
cients can be
2−2 −
1
0.25 − 1
y1
−1
1
y1
y= =
= Ay ,
y2
0 −2
y2
and the eigenvalues of A, −1 and −2, are both negative.
9.4. (a ) Yes, because the eigenvalue, −5, is negative, and hence the solutions
decay exponentially. (b ) No, since 1 + hλ = 1 + 0.5 (−5) = 1.5 > 1. (c )
y1 = 1 + 0.5(−5) = −1.5. (d ) Yes, since backward Euler is unconditionally stable.
(e ) y1 = 1 + 0.5 (−5y1 ) ⇒ y1 = 1/3.5 = 0.2857.
9.5. (a ) y1 = y0 + hf (y0 ) = 1 + 1 · (−1) = 0. (b ) y1 = y0 + hf (y1 ) = 1 + 1 · (−y1 ),
so y = 0.5.
1
9.6. The ﬁxedpoint iteration procedure is
3
yn+1 = 1 − yn /2, with y0 = 1 (or y0 = 0.5 if forward Euler is used to obtain the starting guess).
Note that for the ﬁxedpoint iteration function g (y ) = 1 − y 3 /2, we have g (y ) = 117 Exercises
9.1. (a )
= u2 ,
= t + u1 + u2 . u
1
u
2
(b )
u
1
u
2
u
3 = u2 ,
= u3 ,
= u3 + tu1 . (c )
u
1
u
2
u
3 = u2 ,
= u3 ,
= u3 − 2u2 + u1 − t + 1. 9.2. (a )
u
1
u
2 Exercises 116 = u2 ,
= u2 (1 − u2 ) − u1 .
1 117 (b )
u
1
u
2
u
3 =
=
= u2 ,
u3 ,
−u1 u3 . (c )
u
1
u
2
u
3
u
4 =
=
=
= u2 ,
−GM u1 /(u2 + u2 )3/2 ,
1
3
u4 ,
−GM u3 /(u2 + u2 )3/2 .
1
3 9.3. Yes, since this ﬁrstorder, linear, homogeneous system with constant coeﬃcients can be written
y1
−1
1
y1
y=
=
= Ay ,
y2
0 −2
y2
and the eigenvalues of A, −1 and −2, are both negative. 9.4. (a ) Yes, because the eigenvalue, −5, is negative, and hence the solutions
decay exponentially. (b ) No, since 1 + hλ = 1 + 0.5 (−5) = 1.5 > 1. (c )
y1 = 1 + 0.5(−5) = −1.5. (d ) Yes, since backward Euler is unconditionally stable.
(e ) y1 = 1 + 0.5 (−5y1 ) ⇒ y1 = 1/3.5 = 0.2857.
9.5. (a ) y1 = y0 + hf (y0 ) = 1 + 1 · (−1) = 0. (b ) y1 = y0 + hf (y1 ) = 1 + 1 · (−y1 ),
so y1 = 0.5. ...
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This note was uploaded on 01/22/2012 for the course COT 4501 taught by Professor Davis during the Spring '08 term at University of Florida.
 Spring '08
 Davis

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