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# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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o 2 ^ (_1) m+1sinmnx
Sn = ~ L .
n m
m=1
2b. We have Rn = 11[x _ Sn(x)] 2dx, where Sn(x) is given in
0
part a. Using appropriate computer software we find
Ri = .1307, R2 = .0800, R5 = .0367, R10 = .0193, R15 = .0131
and R19 = .0104.
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 n^/T/enx2/2 and use L'Hopital's Rule:
nv x V x
lim ----- = lim ------ = 0 for x Ô 0. For x = 0,
_nx2/2 v2
e x enx2/2
2
Sn(0) = 0 and thus lim Sn(x) = 0 for all x in [0,1].
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 = 11r(x)[f(x) - S (x) ]2dx
n Q n
n
= J 1r(x)f2(x)dx - 2^^c^ 1r(x)f(x)ôl (x)dx
i=1
nn
+ I Ir(x)ô,(x)ôj(x)dx, i=1 j=1
where the last term is obtained by calculating Sj4x). Using Eqs.(1) and (9) this becomes
nn
R = 1r(x)f2(x)dx-2l c.a. + I c
n Jo i i i
i=1 i=1 nn
= Jo1r(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) > 0) 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 = 0 and
Q 1 Q 1 2
(f ,f ) = 11(c +c x)2dx = 1. Evaluating the integrals
11 Jo 1 2
yields c1 + c2/2 = 0 and c^ + c1c2 + c^/3 = 1, which have
the solution c1 = V"3, c2 = -2\pb and thus
f1(x) = \[3 (1-2x).
7c. f (x) = c + cx + cx2 must satisfy (f ,f ) = 0,
2 1 2 3 0 2
(f1,f2) = 0 and (f2,f2) = 1.
7d. For a (x) = c + ñ x + ñ x2 we have (g„,g„) = 0 and
-'2 1 2 3 30' 32
(g1,g2) = 0, 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 = 0, i Ô j. For
i j -1 i j
i = 0 we have P (x) = 1 and for i = 0 and j = 1 we
0
256
9a.
9b.
9c.
9d.
9e.
10.
12.
Section 11.6
have(P ,P ) = J1 (c,+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 .
³ i
Eq.(6) shows that R > 0 since r(x) > 0 and thus
n
n
J01r(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 = Ó a2 is bounded above. Thus limT exists,
n 4-³ 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 aiô i(x) converges to f(x) in the
i=1
»
mean, then R ^ 0 as n ^ ». Hence ^r(x)f2(x)dx = y a2.
n J0 i
»
Conversely, if J1r(x)f2(x)dx = ai,
i=1
»
alôi(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 ,..., a. = 1/y n ,...,
» »
and Ó a. = Ó 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
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