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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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n = 1,2,... . It can be shown that an arbitrary
piecewise continuous function on [-1,1] can be expressed as a linear combination of Legendre polynomials. Hence we assume that
254
Section 11.6
u(p,$) = z cnpnPn(cos^). The B.C. u(1,^) = f(^) requires
n=1
that u(1,$) = cnPn(cos^) = f(^), 0 < ^ < n. From
n=1
Problem 28 of Section 5.3 we know that P (x) are
n
orthogonal. However here we have P^(cos^) and thus we
must rewrite the equation in Problem 9b to find -[(sin^)O']' = p.2(sin^)0. Thus Pn(cos^) and Pm(cos^) are
orthogonal with weight function sin^. Thus we must multiply the series expansion for f(^) by sin^Pm(cos^)
and integrate from 0 to n to obtain
cm = Jnf(^)sin^Pm(cos^)d^/Jnsin^Pm(cos^)d^. To obtain the answer as given in the text let s = cos$.
Section 11.6, Page 675
2a. bm = l/T J1xsinmnxdx = \p2 (-1) m+1)/mn and thus o 2 ^ (-1) m+1sinmnx
Sn = ~ y. .
n m
m=1
2b. We have Rn = 11[x - Sn(x)] 2dx, where Sn(x) is given in Jo
part a. Using appropriate computer software we find
Ri = .1307, R2 = .0800, R5 = .0367, R10 = .0193, R15 = .0131
and R19 = .o1o4.
2c. As in part b, we find R20 = .0099, and thus n = 20 will insure a mean square error less than .01.
4a. Write S (x) as ni/T/enx2/2 and use L'Hopital's Rule:
nv x V x
lim ---------- = lim = 0 for x ^ 0. For x = 0,
n——? _nx2/2 n——? v2
e x enx2/2
2
Sn(0) = 0 and thus lim Sn(x) = 0 for all x in [0,1].
n—?
Rn > f 1[0-Sn(x)] 2dx = n2 f 1xe-nx2dx 00
n 2 1 n
= - — e |0 = — (1-e ). Since e — 0 as n — ?, we
2 2
have that Rn — ? as n — ?.
255
5. Expanding the integrand we get
R = J 1r(x)[f(x) - S (x) ]2dx
n Jq n
n
= J 1r(x)f2(x)dx - 2^^CiJ 1r(x)f(x)^, (x)dx
i = 1
nn
+ I I cti1 r(x)^,(x) ^j(x)dx, i=1 j=1
where the last term is obtained by calculating S^x). Using Eqs.(1) and (9) this becomes
nn
R = 1r(x)f2(x)dx-2l c . a . + I c
n Jq i i i
i=1 i=1
nn
= JQ1r(x)f2(x)dx - I ai + I (c, - a,)2, by completing i=1 i=1
the square. Since all terms involve a real quantity
squared (and r(x) > Q) we may conclude R^ is minimized by
choosing ci = ai. This can also be shown by calculating
dR /dc. = 2(c.-a.) and setting equal to zero. n i i i
7b. From part a we have fQ(x) = 1 and thus f1(x) = c1 + c2x
must satisfy (f ,f ) = r(c +c x)dx = Q and
Q 1 Jq 1 2
(f ,f ) = 11(c +c x)2dx = 1. Evaluating the integrals 11 Jq 1 2
yields c1 + c2/2 = Q and c^ + c1c2 + c^/3 = 1, which have
the solution c1 = V"3, c2 = -2\f3 and thus
f1(x) = (1-2x).
7c. f (x) = c + cx + cx2 must satisfy (f ,f ) = Q,
2 1 2 3 0 2
(f1,f2) = Q and (f2,f2) = 1.
7d. For a (x) = c + c x + c x2 we have (g„,g„) = Q and
^2 1 2 3 -'2
(g1,g2) = Q, which yield the same ratio of coefficients as found in 7c. Thus g2(x) = cf2(x), where c may now be found from g2(1) = 1.
8. This problem follows the pattern of Problem 7 except now the limits on the orthogonality integral are from -1 to
1. That is (P.,P.) = 1 P. (x)P. (x)dx = Q, i ^ j. For
i j -1 i j
i = Q we have P (x) = 1 and for i = Q and j = 1 we
Q
256
9a.
9b.
9c.
9d.
9e.
10.
12.
Section 11.6
have(P ,P ) = I1 (c1+c_x)dx = (cx+cx2/2)
0 1 -1 1 2 1 2
= 2c1 = 0 and
1
thus P(1) = 1 yields P1(x) = x. The others follow in a similar fashion.
This part has essentially been worked in Problem 5 by setting c = a .
i i
Eq.(6) shows that R > 0 since r(x) > 0 and thus
n
n
|01r(x) f2(x)dx - ai > 0. The result follows. i = 1
Since f is square integrable, r(x)f2(x)dx = M < » and
0
therefore the monotone increasing sequence of partial
n
sums T = X a2 is bounded above. Thus limT exists,
n 4—i i n^» n
i=1
which proves the convergence of the given sum.
This result follows from part a and part c.
»
By definition if a^i(x) converges to f(x) in the
i=1
»
mean, then R ^ 0 as n ^ ». Hence ^r(x)f2(x)dx = X a2.
n J0 i—t i
»
Conversely, if 11r(x)f2(x)dx = ai,
i=1
»
a^i(x) converges to f(x) in the mean.
1 , x a2
i = 1
a2, limR = 0 and
n^» n
i=1
i=1
Bessel's inequality implies that ai converges and thus
i=1
the nth term a ^ 0 as n ^ ».
If the series were the eigenfunction series for a square
»
integrable function, the series ai would have to
i=1
converge. But a0 = 1, a1 = 1/y 2 ,..., an = 1/y n ,...,
» »
and I an = X 1/n is the well-known harmonic series which
n=1 n=1
does not converge.
1
n
ODE ARCHITECT
Companion
ODE ARCHITECT
Companion
CODEE
(Consortium for ODE Experiments)
JOHN WILEY & SONS, INC.
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Copyright 0 1999, by John Wiley & Sons, Inc.
All rights reserved.
Reproduction or translation of any part of this work beyond that permitted by Sections 107 and 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Requests for permission or further information should be addressed to the Permissions Department, John Wiley & Sons, Inc.
ISBN 0-471-
Printed in the United States of America 10 987654321
Printed and bound by
PREFACE
This workbook was designed to accompany the software package ODE Architect, and that’s why we call it a Companion. Each of the 13 Companion chapters corresponds to a multimedia module in the Architect and provides background and opportunities for you to extend the ideas contained in the module. Each chapter ends with several problem sets, called Explorations, related to the chapter and module topics. The Exploration pages can be photocopied so that you can write in answers and derivations, and hand them in along with printouts of graphs produced by the Architect. There is also a notepad facility in the Architect which, with the cut and paste features, makes it possible to write reports.
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