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

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-2 2 nnx 8x nnx 16 nnx

= [-- (2-x )cos------ - sin------ - -cos---]

nn 2 n2n2 2 n3n3 2

4 16

= (1+cosnn) + ---------- (1-cosnn) and thus

nn n3n3

4n2n 2(1+cosnn) + 16(1-cosnn) , nnx

4n n (1+cosnn) + 16(1-cosnn) 1

(x) = V (------------------n----------------)sin-

n3n 3

n=1

2

28b.For the cosine series(even extension) we have

2 |*1 j 1

a0 = xdx =

2 0 2

Section 10.4

215

2 f1 nnx

xcos-----------dx

2 J0 2

2x nnx

[ sin---------- +

nn 2

4 nnx

cos-------]

2 2 n2n 2

2 nn

sin +

nn 2 n2n2

4 nn

cos 2

4

22 n2n 2

so

ro

(x) 4 + ^

n=1

4cos(nn/2) + 2nnsin(nn/2) - 4 nnx

------------------------------------cos----

n2n2 2

For the sine series (odd extension) we have

- 2f1 I

20

2 Γ1 nnx -2x nnx

I xsin--------dx [-------cos------ +

2 ^0 2 nn 2 n2n2

4 nnx

sin------]

2 nn

----cos------ +

nn 2 n2n2

4 nn

sin------

2

so

h(x)

4sin(nn/2) 2nncos(nn/2) nnx

--------------------------------sin-----

n2n2 2

28c.

28d. The maximum error does not approach zero in either case, due to Gibb's phenomenon. Note that the coefficients in both series behave like 1/n as n ^ ro since there is an n in the numerator.

31. We have JLf(x)dx = J

f(x)dx + I Lf(x)dx. Now, if we let

-L J0

y in the first integral on the right, then

f(x)dx

f(x)dx

= I0 = -I

0f(-y)(-dy) = JLf(-y)dy

0

Lf(y)dy + JLf(x)dx

0

= -I³

f(y)dy. Thus

0.

a

n

32. To prove property 2 let f1 and f2 be odd functions and let f(x) = f1(x) ± f2(x). Then f(-x) = f1(-x) ± f2(-x) =

-f1(x) ± [-f2(x)] = -f1(x) + f2(x) = -f(x), so f(x) is

odd. Now let g(x)= f1(x)f2(x), then g(-x) = f1(-x)f2(-x) = [-f1(x)][-f2(x)] = f1(x)f2(x) = g(x)

and thus g(x) is even. Finally, let h(x) = f1(x)/f2(x) and hence h(-x) = f1(-x)/f2(-x) = [-f1(x)]/[-f2(x)] =

f1(x)/f2(x) = h(x), which says h(x) is also even.

Property 3 is proven in a similar manner.

216

Section 10.4

34. Since F(x) = xf(t)dt we have

0

F(-x) = f(t)dt = -I f(-s)ds by letting t = -s. If f

00

is an even function, f(-s) = f(s), we then have

F(-x) = - xf(s)ds = -F(x) from the original definition of 0

F. Thus F(x) is an odd function. The argument is similar if f is odd.

35. Set x = L/2 in Eq.(6) of Section 10.3. Since we know f

is continuous at L/2, we may conclude, by the Fourier

theorem, that the series will converge to

f(L/2) = L at this point. Thus we have

L = L/2 + (2L/n) Σ (-1) n+1/(2n-1), since

n=1

n+1

sin[(2n-1)n/2] = (-1) n+1. Dividing by L and simplifying

-V-, . 1 ^ -V-,

(-1)n+1 V³ (-1)n

4 ' (2n-1) ' 2n+1

n=1 n=0

37a. Multiplying both sides of the equation by f(x) and integrating from 0 to L gives

L[f(x)]2dx = I L[f(x) Σ bnsin(nnx/L)]dx

JoL[f(x)] 2dx = JQL[f(x) Σ1

n=1

= bnJLf(x)sin(nnx/L)dx = (L/2) όΟ, by Eq.(8).

n=1 n=1

This result is identical to that of Problem 17 of Section

10.3 if we set an = 0, n = 0,1,2,... , since

1 L 2 2 L 2

[f(x)]2dx = [f(x)]2dx . In a similar manner, it

L -L L 0

can be shown that

(2/L)|L[f(x) ] 2dx = a2/2 + Σ a°.

n=1

37b. Since f(x) = x and bn = 2L(-1) n+1/nn (from Eq.(9)), we

n

obtain

(2/L)JL[f(x)]2dx = (2/L)JLx2dx = 2L2/3 = Σ^

όΟ =

n=1

Σ4L2/n2n2 = 4L2/n2 Σ(1/n2) or ο2/6 = Σ(1/n2).

n=1 n=1 n=1

Section 10.5

217

We assume that the extensions of f and f' are piecewise continuous on [-2L,2L]. Since f is an odd periodic function of fundamental period 4L it follows from properties 2 and 3 that f(x)cos(nnx/2L) is odd and f(x)sin(nnx/2L) is even. Thus the Fourier coefficients of f are given by Eqs.(8) with L replaced by 2L; that is an = 0, n = 0,1,2,... and

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