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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
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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
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