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

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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 ^ ^.

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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

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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

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ODE ARCHITECT

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CODEE

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