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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 ^ ^ since there is an n in the numerator.
31. We have |L f(x)dx = J
f(x)dx + I Lf(x)dx. Now, if we let
y in the first integral on the right, then
0f(-y)(-dy) = JLf(-y)dy
Lf(y)dy + JLf(x)dx
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.
34. Since F(x) = xf(t)dt we have Jo
F(-x) = f(t)dt = - f(-s)ds by letting t = -s. If f Jo Jo
is an even function, f(-s) = f(s), we then have
F(-x) = - xf(s)ds = -F(x) from the original definition of Jo
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) ^ (-1) n+1/(2n-1), since
sin[(2n-1)n/2] = (-1)n+. Dividing by L and simplifying
-V-, , 1 CO __
(-1) n+1 V1 (-1)n
. n n n V (-1) v
yields — = 7 ------- = 7 -
4 ^(2-1) ^
4 ' (2n-1) ' 2n+1
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) 7 bnsin(nnx/L)]dx
JoL[f(x)] 2dx = |L[f(x) 71
= bnJLf(x)sin(nnx/L)dx = (L/2) bnn, by Eq.(8).
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 + 7an.
37b. Since f(x) = x and bn = 2L(-1) n+1/nn (from Eq.(9)), we
(2/L)JL[f(x)]2dx = (2/L)JLx2dx = 2L2/3 =
74L2/n2n2 = 4L2/n2 7 d/n2) or n2/6 = 7 (1/n2).
n=1 n=1 n=1
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
bn = (2/2L)I2Lf(x)sin(nnx/2L)dx, n = 1,2,... . The
Fourier sine series for f is f(x) =
39. From Problem 38 we have bn = (1/L) 2Lf(x)sin(nnx/2L)dx
= (1/L)ILf(x)sin(nnx/2L)dx + (1/L)I2Lf(2L-x)sin(nnx/2L)dx Jo Jl
= (1/L) ILf(x)sin(nnx/2L)dx - (1/L) I0f(s)sin[nn(2L -s)/2L]ds Jo Jl
and thus bn = 0 for n even and
bn = (2/LH Lf(x)sin(nnx/2L)dx for n odd. The Fourier 0
series for f is given in Problem 38, where the bn are given above.
Section 10.5, Page 579
3. We seek solutions of the form u(x,t) = X(x)T(t).
Substituting into the P.D.E. yields X"T + X'T' + XT' =
X"T + (X' + X)T' = 0. Formally dividing by the quantity
(X' + X)T gives the equation X"/(X' + X) = -T'/T in which
the variables are separated. In order for this equation to be valid on the domain of u it is necessary that both sides be equal to the same constant X. Hence
X"/(X' + X) = -T'/T = X or equivalently,
X" - X(X' + X) = 0 and T' + XT = 0.
5. We seek solutions of the form u(x,y) = X(x)Y(y).
Substituting into the P.D.E. yields X"Y + (x+y)XY" =
X"Y + xXY" + yXY" = 0. Formally dividing by XY yields X"/X + xY"/Y + yY"/Y = 0. From this equation we see that the presence of the independent variable x multiplying the term uyy in the original equation leads to the term xY"/Y when we attempt to separate the variables. It follows that the argument for a separation constant does not carry through and we cannot replace the P.D.E. by two
8. Following the procedures of Eqs.(5) through (8), we set u(x,y) = X(x)T(t) in the P.D.E. to obtain X"T = 4XT', or x"x = 4T'/T, which must be a constant. As stated in the text this separation constant must be -X2 (we choose -X2 so that when a square root is used later, the symbols are simpler) and thus X" + X 2X = 0 and T + (X 2/4)T = 0. Now u(0,t) = X(0)T(t) = 0, for all t > 0, yields X(0) = 0, as discussed after Eq.(11) and similarly u(2,t) = X(2)T(t) = 0, for all t > 0, implies X(2) = 0. The D.E. for X has the solution X(x) = CicosXx + C2sinXx and X(0) = 0 yields Ci = 0. Setting x = 2 in the remaining form of X yields X(2) = C2sin2X = 0, which has the solutions 2X = nn or X = nn/2, n = 1,2,... .
Note that we exclude n = 0 since then X = 0 would yield
X(x) = 0, which is unacceptable. Hence